TJ 230 
M2 
1919 
Copy 1 





















MACHINE DESIGN 


A MANUAL OF PRACTICAL INSTRUCTION IN DESIGNING 
MACHINERY FOR SPECIFIC PURPOSES, INCLUDING 
SPECIFICATIONS FOR BELTS, SCREWS, PINS, 

GEARS, ETC., AND MANY WORKING 
HINTS AS TO OPERATION AND 
CARE OF MACHINES 


ERNEST L. WALLACE, B.S., E.E. 

PATENT ATTORNEY, LOS ANGELES, CALIFORNIA 
FORMERLY ASSISTANT EXAMINER, UNITED STATES 
PATENT OFFICE, WASHINGTON, D. C. 



* 


AMERICAN TECHNICAL SOCIETY 
CHICAGO 

1919 



Copyright, 1918, 1919, 

BY 

AMERICAN TECHNICAL SOCIETY 

Copyrighted in Great Britain 
All Rights Reserved 










©a A 525511 




"Vic I 



INTRODUCTION 


M ACHINE DESIGN is a subject which is often neglected by 
men who have the ambition to graduate from the drafting 
and machine work of our factories into the more important depart¬ 
ments dealing with the design of tools and machines, and the numer¬ 
ous devices which are made by means of them. The draftsman is 
liable to forget that, in order to be a really good draftsman, he 
must add to his ability to draw objects the power of visualizing 
accurately the device or machine, the different views of which he 
is called upon to draw, and be a good enough mechanic to know 
“how things work” and how the machine is to be manufactured. 
Similarly, the machinist is liable to lose sight of the fact that he is 
all the better workman if he can couple with his skill in turning out 
the finished product machined to the proper size, the ability to read 
the draftsman’s drawings down to the smallest detail and to know 
the why and wherefore of every element of the design. 

^ It is plain, therefore, that machine designing in its broadest 
sense demands a familiarity with the point of view of both drafts¬ 
man and machinist; it demands a knowledge of materials—their 
strength, their characteristics, and their behavior under different 
machine operations. The designer must know which parts should 
be cast and which should be machined from steel or brass; he must 
also know the standard designs and specifications for bolts, screws, 
nuts, pins, keys, etc.; he must know the different types of trans¬ 
mission, the proportions of pulleys and gears, the strength of shafts, 
the design of bearings, and the methods of lubrication. 

It is with the idea of offering a simple treatment of these phases 
of machine design that this text was prepared. It is the hope of 
the publishers that the presentation will appeal to young men who 
are anxious to get ahead in their profession as well as to those of 
greater experience in the machine-design field. 



CONTENTS 


PAGE 


Introduction. 

Definition. 

Object. 

Relation. 

Invention. 

Handbooks and empirical data. 

Calculations, notes, and records. 

Strength of materials. 

Stress. 

Unit stress.. 

Unit deformation. 

Experimental laws. 

Elasticity.. 

Ultimate stress.. 

Safe working strength. 

Factor of safety. 

Tension. 

Compression. 

Shear. 

Beams. 

Summation of forces. 

Moment of a force. 

Principle of moments. 

Determination of reactions on beams 

External shear and bending moment. 

Strength of beams. 

Resisting moment of beams. 

Deflection of beams. 

Strength of shafts. 

Shafts. 

Twisting moment. 

Torsional stress. 

Resisting moment. 

Formula for the strength of a shaft. . 

Combined stresses. 

Strength of pipes and cylinders. 


1 

1 

1 

3 

5 

5 

6 

8 

8 

8 

9 

9 

10 

11 

11 

11 

12 

12 

13 

13 

14 

14 

15 

16 
19 
22 
22 

23 

24 
24 
24 
26 
26 

27 

28 
29 


Method of design. 30 

Analysis of conditions and forces. 31 

Theoretical design. 31 

Practical modification.. 32 

Delineation and specification. 34 

Detail drawings. 35 

General drawing. 36 













































CONTENTS 


PAGE 


Lubrication. 37 

Friction. 37 

Lubricant. 38 

Means for lubrication. 38 

Materials employed in construction. 39 

Cast iron. 40 

Wrought iron. 42 

Steel. 43 

Copper. 43 

Aluminum. 44 

Bronze. 44 

Brass. 44 

Fusible alloys. 45 

Bearing alloys. 45 

Wood. 45 

Bolts, studs, nuts, and screws. 45 

Analysis. 46 

Theory. 48 

Practical modification. 52 

Rivets and riveted joints. 57 

Analysis. 57 

Theory. 58 

Practical modification. 59 

Keys and pins. 62 

Analysis. 62 

Theory. 63 

Practical modification. 65 

Cotters. 67 

Analysis. 67 

Theory. 68 

Practical modification. 70 

Couplings. 72 

Analysis. 72 

Theory. 73 

Practical modification. 74 

Friction clutches. 77 

Analysis.*. 78 

Theory. 79 

Practical modification. 81 

Leather belts. v 82 

Analysis. 83 

Theory. 85 

Practical modification. 86 

Strength of leather belting. 88 














































CONTENTS 


Leather belts (continued) page 

Speed of belting. 88 

Material of belting. 89 

Initial tension in belt. 89 

Care of belts. 90 

Methods for joining ends of belt. 91 

Hemp and cotton rope. 93 

Analysis. 93 

Theory. 93 

Practical modification. 94 

Materials. 95 

Systems of rope driving.-. 96 

Wire rope. 97 

Theory. 97 

Practical modification. 97 

Pulleys for wire ropes. 98 

Pulleys. 100 

Analysis. 100 

Theory.:. 102 

Practical modification. 106 

Shafts. Ill 

Analysis. Ill 

Theory. 114 

Practical modification. 121 

Spur Gears. 126 

Analysis. 127 

Theory. 130 

Rim, arms, and hub. 133 

Analysis. 133 

Theory. 133 

Practical modification. 134 

Bevel gears. 138 

Analysis. 138 

Theory. 138 

Practical modification. 139 

Worm and worm gear. 140 

Analysis. 141 

Theory. 141 

Practical modification. 144 

Bearings, brackets, and stands. 146 

Analysis. 147 

Theory. 149 

Practical modification. 154 













































SECTIONAL ELEVATION OF NEW CROSSLEY OIL ENGINE 

Courtesy of Crossley Brothers, Manchester, England 









































































































































\ 




MACHINE DESIGN 

PART I 


INTRODUCTION 

Definition. Machine Design is the art of mechanical thought, 
development, and specification. 

It is an art, in that its routine processes may be analyzed and 
systematically applied. Proficiency in the art positively cannot 
be attained by any “short cut” method. There is nothing of a 
spectacular nature in the methods of Machine Design. Large 
results cannot be accomplished at a single bound, and success is 
possible only by a patient, step-by-step advance in accordance with 
well-established principles. 

Mechanical thought means the thinking of things strictly from 
their mechanical side; a study of their mechanical theory, structure, 
production, and use; a consideration of their mechanical fitness as 
parts of a machine. 

Mechanical development signifies the taking of an idea in the 
rough—in the crude form, for example, in which it comes from the 
inventor—working it out in detail, and refining and fixing it in shape 
by the designing process. Ideas in this way may become com¬ 
mercially practicable designs. 

Mechanical specification implies the detailed description of de¬ 
signs in such exact form that the shop workmen are enabled to con¬ 
struct completely and to put in operation the machines represented 
in the designs. 

Object. The object of Machine Design is the creation of ma¬ 
chinery for specific purposes. Every department of a manufacturing 
plant is a controlling factor in the design and production of the 
machines built there. A successful design cannot be out of harmony 
with the organized methods of production. Hence, in the high 
development of the art of Machine Design is involved a knowledge 



2 


MACHINE DESIGN 


of the operations in all the departments of a manufacturing plant. 
The student is therefore urged not only to familiarize himself with 
the direct production of machinery, but to study the relation thereto 
of the allied commercial departments. He should get into the spirit 
of business at the start, get into the shop atmosphere, execute his 
work just as though the resulting design were to be built and sold 
in competition. He should visit shops, work in them, if possible, 
and observe details of design and methods of finishing machine 
parts. In this way he will begin to store up bits of information, 
practical and commercial, which will have valuable bearing on his 
engineering study. 

The labor involved in the design of a complicated automatic 
machine is evidenced by the designer’s wonderful familiarity with 
its every detail as he stands before the completed machine in opera¬ 
tion and explains its movements to an observer. The intricate 
mass of levers, shafts, pulleys, gears, cams, clutches, etc., packed 
into a small space, and confusing even to a mechanical mind, seems 
like a printed book to the designer of them. 

This is so because it is a familiar journey for the designer’s 
mind to run over a path which it has already traversed so many 
times that he can see every inch of it with his eyes shut. Every 
detail of that machine has been picked from a score or more of pos¬ 
sible ideas. One by one, ideas have been worked out, laid aside, 
and others taken up. Little by little, the special fitness of certain 
devices has become established, but only by patient, careful con¬ 
sideration of others, which, at first, seemed equally good. 

Every line, and corner, and surface of each piece, however 
small that piece may be, has been through the refining process of 
theoretical, practical, and commercial design. Every piece has 
been followed in the mind’s eye of its designer from the crude material 
of which it is made, through the various processes of finishing, to its 
final location in the completed machine; thus its bodily existence 
at that point is but the realization of an old and familiar picture. 

What wonder that the machine seems simple to the designer 
of it! As he looks back to the multitude of ideas invented, worked 
out, considered, and discarded, the machine in its final form is but 
a trifle. It merely represents a survival of the fittest. 

No successful machine, however simple, was ever designed 


MACHINE DESIGN 


3 


that did not go through this slow process of evolution. No machine 
ever just simply happened by accident to do the work for which it 
is valued. No other principle upon which the successful design 
of machinery depends is so important as this careful, patient con¬ 
sideration of detail. A machine is seldom unsuccessful because 
some main point of construction is wrong. The principal features 
of a machine are usually the easiest to determine. It is a failure 
because some little detail was overlooked, or hastily considered, 
or allowed to be neglected, because of the irksome labor necessary 
to work it out properly. 

There is no task so tedious, for example, as the devising of the 
method of lubricating the parts of a complicated machine. Yet 
there is no point of design so vital to its life and operation as an ab¬ 
solute assurance of an adequate supply of oil for the moving parts 
at all times and under all circumstances. Suitable means often 
cannot be found, after the parts are together, hence the machine 
goes into service on a risky basis, with the result, perhaps, of early 
failure, due to “running dry.” Good designers will not permit a 
design to leave their hands which does not provide practically auto¬ 
matic oiling, or at least such means of lubrication that the operator 
can offer no excuse for neglecting to oil his machine. This is but 
a single illustration of many which might be presented to impress 
the definite and detail character necessary in work in Machine 
Design. 

Relation. The relation which Machine Design should cor¬ 
rectly bear to the problems that it seeks to solve, is twofold; and 
there are, likewise, two points of view corresponding to this two¬ 
fold relation, from which a study of the subject should be traced, 
viz, theory and production. Neither of these can be discarded and 
an efficient mastery of the art attained. 

Theory. From this point of view, Machine Design is merely 
a skeleton or framework process, resulting in a representation of 
ideas of pure motion, fundamental shape, and ideal proportion. 
It implies a working knowledge of physical and mathematical laws. 
It is a strictly scientific solution of the problem at hand, and may 
be based purely on theory which has been reasoned out by calculation 
or deduced from experiment. This is the only sure foundation for 
intelligent design of any sort. 


4 


MACHINE DESIGN 


But it is not enough to view the subject from the standpoint 
of theory alone. If a stop were made here nothing would be obtained 
but mechanisms, mere laboratory machines, simply structures of 
ingenuity and examples of fine mechanical skill. A machine may 
be correct in the theory of its motions; it may be correct in the theoreti¬ 
cal proportions of its parts; it may even be correct in its operation 
for the time being; and yet its complication, its misdirected and 
wasteful effort, its lack of adjustment, its expensive and irregular 
construction, its lack of compactness, its difficulty of ready repair, 
its inability to hold its own in competition—any of these may throw 
the balance to the side of failure. Such a machine, commercially 
considered, is of little value. No shop will build it, no machinery 
house will sell it, nobody will buy it if it is put on the market. Thus, 
aside from the theoretical correctness of principle, the design of a 
machine must satisfy certain other exacting requirements of a dis¬ 
tinctly business nature. 

Production. From this point of view, Machine Design is the 
practical, marketable development of mechanical ideas. Viewed 
thus, the theoretical, skeleton design must be clothed and shaped 
that its production may be cheap, involving simple and efficient proc¬ 
esses of manufacture. It must be judged by the latest shop methods 
for exact and maximum output. It must possess all the good points 
of its competitor, and, withal, some novel and valuable ones of its 
own. In these days of keen competition it is only by carefully 
studied, well-directed effort toward rapid, efficient, and, therefore, 
cheap production that any machine can be brought to a commercial 
basis, no matter what its other merits may be. All this must be 
thought of and planned for in the design, and the final shapes arrived 
at are quite as much a result of this second point of view as of the first. 

As a good illustration of this, may be cited the effect of the present 
somewhat remarkable development of the so-called high-speed 
steels. The speeds and feeds possible with tools made of these 
steels are such that the driving power, gearing, and feed mechanism 
of the ordinary lathe are wholly inadequate to the demands made 
upon them when working the tool to its limit. This means that the 
basis of design as used for the ordinary tool steel will not do if the 
machine is expected to stand up to the cuts possible with the new 
steels. Hence, while the old designs were right for the old standard, 


MACHINE DESIGN 


5 


a new one has been set, and a thorough revision on a high-speed basis 
is imminent, else the market for them as machines of maximum out¬ 
put will be lost. 

It is evident, therefore, that the designer must not only use all 
the theory at his command, but must continually inform himself 
on all processes and conditions of manufacture, and keep an eye on 
the tendency of the sales markets, both of raw material and the 
finished machinery product. Thought which is directed and con¬ 
trolled, not only by theoretical principle but by closely observed 
practice, is what, in the broadest sense, is meant by the term mechanical 
thought. From the feeblest pretenders of design to those engineers 
who consummate the boldest feats and control the largest enterprises, 
the process which produces results is always the same. Although 
experience is necessary for the best mechanical judgment, yet the 
student must at least begin to cultivate good mechanical sense very 
early in his study of design. 

Invention. Invention is closely related to Machine Design, 
but is not design itself. Whatever is invented has yet to be designed 
and is of little value until it has been refined by the process of design. 

Although original design is of an inventive nature, it is not 
strictly invention. Invention is the product of genius and though 
announced in a flash of brilliancy is the result of a long course of the 
most concentrated brain effort. It is not spontaneous; it is not thrown 
off like sparks from the blacksmith’s anvil; but it is the result of hard 
and applied thinking. This is worth noting carefully, for the same 
effort which produces original design may develop a valuable inven¬ 
tion. There is little possibility, however, of inventing anything 
except through exhaustive analysis and a clear interpretation of such 
analysis. 

Handbooks and Empirical Data. The subject matter in these 
is often contradictory in its nature, but valuable nevertheless. Em¬ 
pirical data is data for certain fixed conditions and is not general. 
Hence, when handbook data is applied to some specific case of 
design, while the information should be used in the freest manner, 
yet it must not be forgotten that the case at hand is probably dif¬ 
ferent, in some degree, from that upon which the data was based, 
and unlike any other case which ever existed or will ever again exist. 
Therefore the data should be applied with the greatest discretion, 


0 


MACHINE DESIGN 


and when so applied will contribute to the success of the design at 
least as a check, if not as a positive factor. 

The student should at the outset purchase one good handbook, 
and acquire the habit of consulting it on all occasions, checking 
and comparing his own calculations and designs therefrom. Care 
must be taken not to become tied to a handbook to such an extent 
that one’s own results are wholly subordinated to it. Independence 
in design must be cultivated and one should not sacrifice his calculated 
results until they can be shown to be false or based on false assump¬ 
tion. Originality and confidence in design will be the result if this 
course be honestly pursued. 

Calculations, Notes, and Records. Accurate calculations are 
the basis of correct proportions of machine parts. There is a right 
way to make calculations and a wrong way, and the student will 
usually take the wrong way unless he is cautioned at the start. 

The wrong way of making calculations is the loose and shiftless 
fashion of scratching upon a scrap of detached paper marks and 
figures, arranged in haphazard form, and disconnected and in¬ 
complete. These calculations are in a few moments’ time totally 
meaningless, even to the author of them himself, and are so easily 
lost or mislaid that when wanted they usually cannot be found. 

Engineering calculations should always be made systemat¬ 
ically, neatly, and in perfectly legible form, in some permanently 
bound blank book, so that reference may always be had to them at 
any future time for the purpose of checking or reviewing. Put all 
the data down. Do not leave in doubt the exact conditions under 
which the calculations were made. Note the date of calculation. 

If a mistake in figures is made, or a change is found necessary, 
never rub out the figures or tear out the leaf, or in any way obliterate 
the figures. Simply draw a bold cross through the wrong part and 
begin again. Often a calculation which is supposed to be wrong 
is later shown to be right, or the fact which caused the error may 
be needed for investigation and comparison. Time which is spent 
in making figures is always valuable time, time too precious to be 
thrown away by destroying the record. 

The recording of calculations in a permanent form, as just 
described, is the general practice in all modern engineering offices. 
This plan has been established purely as a business policy. In 


MACHINE DESIGN 


7 


case of error it locates responsibility and settles dispute. Con¬ 
sistent designing is made possible through the records of past designs. 
Proposals, estimates, and bids may often be made instantly, on the 
basis of what these record books show of sizes and weights. This 

o 

bookkeeping of calculations is as important a factor of systematic 
engineering as bookkeeping of business accounts is of financial 
success. 

The student should procure for this purpose a good blank book 
with a firm binding, size of page not smaller than 6" X 8"—perhaps 
^XlT' may be better—and every calculation, however small and 
apparently unimportant, should be made in it. 

The development of a personal notebook is of great value to 
the designer of machinery. The facts of observation and experience 
recorded in proper form, bearing the imprint of intimate personal 
contact with the points recorded, cannot be equaled in value by 
those of any hand or reference book made by another. There is 
always a flavor about a personal notebook, a sort of guarantee, 
which makes the use of it by its author definite and sure. 

The habit of taking and recording notes, or even knowing 
what notes to take, is an art in itself, and the student should begin 
early to make his notebook. Aside from the value of the notes 
themselves as a part of his personal equipment, the facility with 
which his eye will be trained to see and record mechanical things 
will be of great value in all of his study and work. How many men 
go through a shop and really see nothing of the operations going 
on therein, or, seeing them, remember nothing! An engineer, 
trained in this respect, will, to a surprising degree, be able to retain 
and sketch little details which fall under his eye for a brief moment 
only, while he is passing through a crowded shop. 

Some draftsmen have the habit of copying all the standard 
tables of the various offices in which they work. While these are 
of some value in a few cases, yet this is not what is meant by a good 
notebook in the best sense. Ideas make a good notebook, not a 
mere tabulation of figures. If the basis upon which standards are 
founded can be transferred to permanent personal record, or novel 
methods of calculation, or simple features of construction, or data 
of mechanical tests, or efficient arrangement of machinery—if these 
can be preserved for reference, the notebook will be of greatest value. 


8 


MACHINE DESIGN 


Whatever is noted down, make clear and intelligible, illustrat¬ 
ing by a sketch if possible. Make the note so clear that Terence 
to it after a long space of years will bring the whole subject'before 
the mind in an instant. If this is not done the author of the note 
himself will not have patience to dig out the meaning when it is 
needed; and the note will be of no value. 

STRENGTH OF MATERIALS 

In order to determine the sizes of members and the arrange¬ 
ment of parts of a machine it is necessary to know the character and 
the effect of the loads applied. 

STRESS 

If a force is applied to any body it tends to change the shape or 
volume of that body. Thus, if a bar of iron is fixed firmly at its 
upper end and a heavy weight is hung from the lower, the bar lengthens. 
If a weight is placed on the top of a block of wood or metal, the block 
is shortened. In the above examples, the weight or force called the 
load tends to change the shape or volume or both, and the forces 
within the body that tend to resist the change are called stresses. The 
word stress is also used as a name for the force. Engineers say that 
a body is strained or stressed when a load is placed upon it. 

Stresses are measured in pounds, tons, grams, kilograms, etc. 
For example, if a weight of fifteen hundred pounds is suspended by 
a rope, the stress in the rope is fifteen hundred pounds. If a block of 
iron has a weight or force of forty tons on its top, the stress is 
forty tons. If the force tends to pull the particles of the body apart, 
it is called tensile stress; if it tends to crush the body, it is called 
compressive stress; and the stress that resists the slipping of one 
section of a body past another section is called a shearing stress. 

Unit Stress. Unit strain or stress is the amount of stress on a 
unit area of section and is expressed as pounds per square inch, tons 
per square inch or per square foot, kilograms per square centimeter, 

etc. Let A equal the area in square inches and P the total stress 

P 

in pounds; then the unit stress will be —. If S equals unit stress, 

A 

P P 

—r = 8, P = AS, A =~7T 
A S 


(i) 


MACHINE DESIGN 


9 


Suppose a brick in a testing machine to crush at a stress of 10 
tons, the section of the brick being 2X4 inches. The unit com¬ 
pressive strain is 


P _ 20,000 
A 8 


2,500 pounds 


If a rope supports a weight of 314.16 pounds, the unit tensile strain 
being 400 pounds per square inch, what is the size of the rope? 


P _ 314.16 
8 400 


.7854 square inches 


Since the area is .7854 square inches, the diameter is 1 inch. 

When a force or weight is applied to a body there is a change 
of form or volume. This change of shape is called a strain or deforma¬ 
tion and may be measured in inches, centimeters, etc. 

Unit Deformation. Unit deformation is the amount of lengthen¬ 
ing or shortening per unit of length. Let L equal length in feet, 
b the elongation or shortening per unit of length, and B the total 
deformation. Then evidently L X b — B, or B = bL. If a bar 
of iron 10 feet long lengthens .1 of an inch under a load of 20,000 
pounds, the elongation per foot will be 

b =-?-= — = .01 inches (2) 

L 10 w 


Experimental Laws. By means of experiments, together with 
experience, the following general laws have been established and may 
be taken as fundamental principles. 

1. When a small stress is applied to a body, a small deformation 
is produced and, on removal of the stress, the body returns to its original 
form. Materials may be regarded as perfectly elastic for small 
stresses. 

2. Under small stresses the deformations are nearly proportional 
to those stresses, and also approximately proportional to the length oj 
the body. 

3. When the stress is sufficiently great, a deformation is pro¬ 
duced which is partly permanent; that is, the body does not spring 
back to its original shape when the stress or strain is removed. In 
such cases, the elastic limit is said to have been exceeded and deforma¬ 
tion is no longer proportional to stress. 

4. If the stress is still further increased, deformation increases 

until the piece breaks. 








10 


MACHINE DESIGN 


5. A sudden stress or shock is more injurious than a steady stress 
gradually applied. 

The words small and great in the above laws have different 
values for different materials; i. e., a large stress for wood would not 
be a large stress for steel. 

Elasticity. According to the first law given above materials 
resume their original form after the stress is removed, provided that 
stress has not been too great. This resistance offered by a body to 
permanent change in form is called elasticity. When the load is 
large and the piece does not return to its original shape, it is said to 
have a permanent set. The unit stress at which permanent set is 
first visible is called the elastic limit. 

The body being perfectly elastic within the elastic limit, laws 
describing its action can be formulated, but beyond the elastic limit, 
there being a permanent alteration of shape, these laws cannot be 
applied. 

In testing a specimen of any material, it is easy to reduce its 


stress to unit stress by the formula S 


P 

A* 


and its deformation to unit 


deformation by the formula 6 =-—. Suppose the unit stress and the 

Jj 

unit deformation are known, then the ratio of these, called the co¬ 
efficient of elasticity , may easily be found. This coefficient expressed 
algebraically is 

T = E (3) 


Examples. 1 . Suppose the unit stress in a specimen under 
test is 45,000 pounds and the unit deformation is .0015 inch. What is E? 
Solution. 


S _ 45,000 
b .0015 


30,000,000 


2. A flat cast-iron foundation ring, 4 inches high, whose area 
is 4 square feet, has a weight of 144 tons placed on the top. If the 
weight causes a shortening of .00016 of an inch, what is the coefficient 
of elasticity? 

Solution. 


S 


P 


B 








MACHINE DESIGN 


11 


Then substituting 

R _ PL _ 144 X 2000 X 4 
AB 4 X 144 X .00016 


12,500,000 


Ultimate Stress. When a bar is under stress exceeding its 
elastic limit, it is usually unsafe. As the load, or stress, is increased, 
the deformation is increased rapidly, and finally the bar ruptures. 
The ultimate strength of the bar is the unit strength which the bar 
offers just before rupture. Sometimes the strength shown by the 
bar just at the time when rupture occurs is less than that shown just 
before breaking, the latter being called the breaking strength. 

Safe Working Strength. For machines and other structures, 
it would be unsafe to exert upon the bodies used an external force 
or load equal to their ultimate strength. It is not even advisable 
to allow the point to be reached at which permanent set occurs. 
The unit stress under which the bodv or material is to act or work, 
is called the working strength. This allowable working strength 
depends upon the character of the material and the load. It is usual 
to divide the ultimate strength by some number to determine the 
working strength. If the ultimate strength of steel is 60,000 pounds, 


the working strength may be 


60,000 

10 


6 , 000 . 


Factor of Safety. The number by which the ultimate strength 
is divided to determine the working strength is called the factor of 
safety. Materials which are unreliable as to their ultimate strength, 
such as stone and cast iron, have a large factor of safety. Materials 
to which a steady load is applied do not require as high a factor of 


TABLE I 


Factor of Safety 


Material. 

For Steady 
Load 

For Varying 
Load 

For Shock 

Cast Iron 

7 

10 

20 

Wrought Iron 

4 

6 

11 

Steel 

5 

7 

12 

Soft Metals and Alloys 

6 

8 

15 

Ropes 

7 

10 

12 

Leather 

9 

12 

15 

Timber 

9 

12 

15 

Brick and Stone 

12 

20 

30 

















12 


MACHINE DESIGN 


l 


safety as those which carry a moving load. The proper factor of 
safety can only be determined by good engineering judgment and 
experience. Average values for materials in extensive use are given 
in Table I. 

Testing Machines. Materials are tested for strength and elas¬ 
ticity by placing specimens of standard shapes in machines designed 
for this purpose, and subjecting them to tension, compression, and 
bending. The stress and the resulting change in the material is 
shown by some indicating devices. 

Tension. Tensile stress in a body is the resistance offered by 
its molecules to being pulled apart. The action of the forces may be 

investigated by subjecting a rod to a heavy down¬ 
ward pull until the molecular force is overcome. 

Suppose the point where the rod may rupture 
is at the cross-section C, Fig. 1. At this point the 
downward pull of 1,000 pounds due to the load is 
resisted by the attraction for each other of the mole¬ 
cules about the section, each downward pull call¬ 
ing forth an equal upward pull from the molecules. 
At any other point along the rod a similar balanced 
condition may be found. Only when an increase 
in the load produces a downward pull which is 
greater than the maximum attractive force of the 
molecules, will a rupture occur. 

If the load is P, the cross-sectional area A , and 
the safe tensile working stress S, then 


) /OOP Lbs. \ 



Fig. 1. Rod in 
Tension 


A = 


S 


Compression. The compressive or crushing strength of any 
material is the resistance offered by its molecules to being pushed 
nearer together. It is the opposite of tension. Fig. 2 represents 
a post carrying a load at its upper end. Consider any section as C. 
The load above this section tends to cause the molecules at the 
section to be pressed closer together, in this case assisting the molec¬ 
ular attraction along the vertical axis. The attraction of the mole¬ 
cules in all other directions, however, prevents the post from being 






















MACHINE DESIGN 


13 


crushed by the load. When the load becomes too great, the molec¬ 
ular resistance is overcome and the post breaks. Loaded posts, 
or struts, piers, etc., are under compressive 
stress. 

In compression, the formula 




LOAD 



- C 



also applies, A being the cross-sectional area, 

P the load applied, and S the safe compressive 
strength. 

Shear. The shearing strength of a ma¬ 
terial is the resistance offered by its fibers to 
being cut, or, if not fibrous, the resistance 
offered by its molecules to being slipped by 
each other. Shear is somewhat different from 
tension and compression; it occurs when 
two forces tend to cut a body between them. 

Let Fig. 3 represent a riveted joint and consider the section 
through the rivet at C as dividing it into two parts, A and B. The 

forces applied to the joint are 


such that A tends to slide to 
the left and B to the right; 
then the molecules on one 
side of section C exert an 


Fig. 2. Post in Com¬ 
pression 



Fig. 3. Rivet in Shear 


attraction on the molecules on the other side of the section which 
tends to prevent slipping. The force which produces the stress in 
the rivet is called a shearing stress. 

As in tension, the cross-sectional area A is found by the formula 



where the symbol P represents the load and S, the safe compressive 
strength. 

Beams. A bar supported in a horizontal position is called a 
beam. A cantilever beam is one resting on one support or fixed at 
one end, as in a wall, the other end being free. A simple beam is one 
resting on two supports. 














14 


MACHINE DESIGN 


Summation of Forces. The sum of all the forces acting in a 
given direction upon a body whose position relative to other bodies 
remains fixed, must be equal to the sum of all the forces acting in 
the opposite direction. If the forces acting in one direction are greater 
than those acting in the opposite direction, motion will result. 

Moment of a Force. By moment of a force with respect to a 
point is meant its tendency to produce rotation about that point. 
Evidently the tendency depends on the magnitude of the force and 
on the perpendicular distance of the line of action of the force from 
the point; the greater the force and the perpendicular distance, the 


greater the tendency; hence the moment of a force with respect to a 

point equals the product of the 
force and the perpendicular dis¬ 
tance from the force to the point. 

The point with respect to 
which the moment of one or 
more forces is taken is called an 
origin or center of moments , and 
the perpendicular distance from 
an origin of moments to the line 
of action of a force is called the 
arm of the force -with respect to 
that origin. Thus, if F x and F 2 , 
Fig. 4, are forces, their arms 
with respect to 0 ' are af and af respectively, and their moments 
are F x a x f and F 2 af. With respect to 0" their arms are a x and a 2 " 
respectively, and their moments are F x a" and F 2 a 2 . 

If the force is expressed in pounds and its arm in feet, the moment 
is in foot-pounds; if the force is expressed in pounds and its arm in 
inches, the moment is in inch-pounds. 

A sign is given to the moment of a force for convenience. The 
rule used herein is as follows: The moment of a force about a point 
is positive or negative according as it tends to turn the body about that 
point in the *clockwise or counter-clockwise direction. Thus the 
moment, Fig. 4, of F x about 0' is negative, about 0", positive; of 
F 2 about 0' is negative, about 0", negative. 



*By clockwise direction is meant that in which the hands of a clock rotate: and by 
counter-clockwise, the opposite direction. 




MACHINE DESIGN 


15 


Principle of Moments. If a line or point in a body is taken as 
an axis about which the body may be considered as tending to rotate, 
then the sum of the moments tending to produce rotation about that 
axis in one direction will be equal to the sum of the moments tending 
to produce rotation in the opposite direction if the body is at rest with 
respect to other bodies. If it is not at rest, the sum of the moments 
in opposite directions is not equal. In general, a single force of 
proper magnitude and direction can balance a number of forces. 

All the forces acting upon a body which is at rest are said to 
be balanced or in equilibrium . No force is required to balance such 
forces and hence their equilibrant and resultant are zero. Since 
their resultant is zero, the algebraic sum of the moments of any num- 
ber of forces which are balanced or in equilibrium equals zero. 

This is known as the principle of moments for forces in equi¬ 
librium; or, briefly, the principle of moments . 


looollos. 

-4- 


aooolbs. 


i 

©s. 


12 * * 


aooolbs. 
at-a'-A 


aooolbs. 




f 


- 4 - r - 

I 


* 


30oolbs. 3 ooolbs. 

Fig. 5. Moments Acting in Beam 


Example. Let AB , Fig. 5, be a beam resting on supports at 
C and F. Find the moments with respect to the supports. 

Solution. It is evident from the symmetry of the loading that 
each reaction equals one-half of the whole load; i. e., J of 6,000 = 
3,000 pounds. (The weight of the beam is neglected for simplicity.) 

With respect to C, for example, the moments of the forces are, 
taking them in order from the left: 


-1,000 X 4 = 
3,000 X 0 = 
2,000 X 2 = 
2,000 X 14 = 
-3,000 X 16 = 
1,000 X 20 = 


— 4,000 foot-pounds 

0 

4,000 

28,000 

-48,000 

20,000 


The algebraic sum of these moments is seen to equal zero. 













16 


MACHINE DESIGN 


Again, with respect to B, the moments are: 

— 1,000 X 24 = —24,000 foot-pounds 
3,000 X 20 = 60,000 

-2,000 X 18 = -36,000 
-2,000 X 6 =-12,000 
3,000 X 4 = 12,000 

1,000 X 0 = 0 

The sum of these moments also equals zero. In fact, no matter 
where the center of moments is taken, it will be found in this and 
any other balanced system of forces that the algebraic sum of their 
moments equals zero. The chief use to be made of this principle 
will be in finding the supporting forces of loaded beams. 

Determination of Reactions on Beams. The forces which the 
supports exert on a beam, i. e., the “supporting forces,” are called 
reactions , and will be considered chiefly in connection with simple 
beams. The reaction on a cantilever beam evidently equals the 
total load on the beam. 

When the loads on a horizontal beam are all vertical—and this 
is the usual case—the supporting forces are also vertical and 
the sum of the reactions equals the sum of the loads. This principle 
is sometimes useful in determining reactions, but in the case of simple 
beams the principle of moments is sufficient. The general method 
of determining reactions is as follows: 

1. Write out two equations of moments for all the forces— 
loads and reactions—acting on the beam with origins of moments 
at the supports. 

2. Solve the equations for the reactions. 

3. As a check, try if the sum of the reactions equals the sum 
of the loads. 


1003 lbs. 


aooo lbs. 


k-1- 


Aar 


5 *- 


3000 ]bs. 


■ 2 1 * 




1 


R»- 


Fig. 6. Reactions at Supports of Simple Beam 


Examples. 1 . Fig. 6 represents a beam supported at its ends 
and sustaining three loads. Find the reactions due to these loads. 











MACHINE DESIGN 


17 


Solution. Let the reactions be denoted by R x and R v as shown; 
then the moment equations are: 

For origin at A 

(1,000 X 1) + (2,000 X 6) + (3,000 X 8) -(R 2 X 10) = 0 
For origin at E 

(R, X 10) - (1,000 X 9) - (2,000 X 4) - (3,000 X 2) = 0 
The first equation reduces to 

10 R 2 = 1,000 + 12,000 + 24,000 = 37,000; or 
R 2 = 3,700 pounds 

The second equation reduces to 

10 R l = 9,000 + 8,000 + 6,000 = 23,000; or 
R t = 2,300 pounds 

The sum of the loads is 6,000 pounds and the sum of the reactions 
is the same; hence the computation is correct. 

aioolbs. 36oo lbs. leoolba.' 


1, V, 





1 -' 

^, 

^ 

D a 

,E 


► 3 



Fig. 7. Reactions in Beam with Overhanging Ends 

2. Fig. 7 represents a beam supported at B and D — i. e., it 
has overhanging ends—and sustaining three loads, as shown. De¬ 
termine the reactions due to the loads. 

Solution. Let R x and R 2 denote the reactions as shown; then 
the moment equations are: 

For origin at B 

(-2,100 X 2) + 0 + (3,600 X 6) - ( R 2 X 14) + (1,600 X 18) =0 * 
For origin at D 

(-2,100 X 16) + (R,X 14) - (3,600 X 8) + 0 + (1,600 X 4) = 0 

The first equation reduces to 

14 R 2 = -4,200 + 21,600 + 28,800 = 46,200; or 
R 2 = 3,300 pounds 

The second equation reduces to 

14 R, = 33,600 + 28,800 - 6,400 = 56,000; or 
R x = 4,000 pounds 











18 


MACHINE DESIGN 


The sum of the loads equals 7,300 pounds and the sum of the reactions 
is the same; hence the computation checks. 

3. What are the total reactions in Example 1, if the beam 
weighs 400 pounds? 

Solution . (1) Since the reactions due to the loads are already 

known, being 2,300 and 3,700 pounds at the left and right ends 
respectively, Example 1, it is only necessary to compute the reactions 
due to the weight of the beam and add. Evidently the reactions 
due to the weight equal 200 pounds each; hence the 

left reaction = 2,300 + 200 = 2,500 pounds 

and the 

right reaction = 3,700 + 200 = 3,900 pounds. 

(2) Or, the reactions due to the loads and weight of the beam 
might be computed together and directly. In figuring the moment 
due to the weight of the beam, or any uniformly distributed load, 
the weight is considered as concentrated at the middle of the beam; then 
its moments with respect to the left and right supports are (400 X 5) 
and —(400 X 5), respectively. The moment equations for origins 
at A and E are like those of Example 1 except that they contain one 
more term, the moment due to the weight; thus they are respectively 

(1,000 X1) + (2,000 X 6) + (3,000 X 8) - (R 2 X10) + (400X 5) = 0 
(R t X10) - (1,000 X 9) - (2,000 X 4) - (3,000 X 2) - (400 X 5) = 0 
The first equation reduces to 

10 R 2 = 39,000, or R 2 = 3,900 pounds 
The second equation reduces to 

107? 1 = 25,000, or R x = 2,500 pounds 

4. What are the total reactions in Example 2, if the beam weighs 
42 pounds per foot? 

Solution. As in Example 3, the reactions due to the weight 
might be computed and then added to the corresponding reactions 
due to the loads (already found in Example 2), but in the following 
solution the total reactions due to load and weight are determined 
directly. 

The beam being 20 feet long, its weight is 42 X 20, or 840 pounds. 
Since the middle of the beam is 8 feet from the left and 6 feet from 
the right support, the moments of the weight with respect to the left 
and right supports are, respectively; 


MACHINE DESIGN 


19 


840 X 8 = 6,720, and —840 X 6 = —5,040 foot-pounds 
The moment equations for all the forces applied to the beam for 
origins at B and D are like those in Example 2, with an additional 
term,, the moment of the weight. They are, respectively: 

(-2,100 X 2) + 0 + (3,600 X 6) - {R 2 X 14) + (1,600 X 18) + 6,720=0 
(-2,100 X 16) + (R t X 14) - (3,600 X 8) + 0 + (1,600 X 4) -5,040 = 0 

The first equation reduces to 

14 R 2 = 52,920, or R 2 = 3,780 pounds 
The second equation reduces to 

14 R t = 61,040, or R x = 4,360 pounds 

The sum of the loads and weight of beam is 8,140 pounds; and since 
the sum of the reactions is the same, the computation checks. 

EXTERNAL SHEAR AND BENDING MOMENT 

On almost every cross-section of a loaded beam the three kinds 
of stress, viz, tension, compression, and shear, appear. Tension and 
compression are often called fiber stresses because they act along the 
real fibers of a wooden beam or the imaginary ones of which it may 
be supposed iron and steel beams are composed. Before taking 
up the subject of these stresses in beams it is desirable to study cer¬ 
tain quantities relating to the loads, and on which the stresses in a 
beam depend. These quantities are called external shear and bend¬ 
ing moment. 

External Shear. By external shear at (or for) any section of 
a loaded beam is meant the algebraic sum of all the loads (includ¬ 
ing weight of beam) and reactions on either side of the section. 
This sum is called external shear because, as is shown later, it equals 
the shearing stress (internal) at the section. For brevity, the term 
“shear” will be used in this discussion when external shear is meant. 

Rule of Signs. In computing external shears, it is customary 
to give the plus sign to the reactions and the minus sign to the loads. 
When the external shear is computed from the loads and reactions 
to the right the sign of the sum is changed in order to get the same 
sign for the external shear, as when computed from the left. Thus 
for section a of the beam in Fig. 5 the algebraic sum when com¬ 
puted from the left 


20 


MACHINE DESIGN 


= -1,000 + 3,000 
= 2,000 pounds 

and when computed from the right 

= -1,000 + 3,000 - 2,000 - 2,000 
= — 2,000 pounds 

The external shear at section a is +2,000 pounds. 

Again, for section b the algebraic sum when computed from the left 

= -1,000 + 3,000 - 2,000 - 2,000+ 3,000 
= +1,000 pounds 

and when computed from the right 

= —1,000 pounds 

The external shear at the section is +1,000 pounds. 

It is usually convenient to compute the shear at a section from 
the forces to the right or left according as there are fewer forces— 
loads and reactions—on the right or left sides of the section. 

Units. It is customary to express external shears in pounds, 
but any other unit for expressing force and weights—as the ton— 
may be used. 

Maximum Shear. It is sometimes desirable to know the greatest 
or maximum value of the shear in a given case. The maximum 
shear is given in Table II for the usual cases of loading. 

In cantilevers fixed in a wall , the maximum shear occurs at the 

wall. 

In simple beams, the maximum shear occurs at a section next to 
one of the supports. 

Bending Moment. By bending moment at (or for) a section of 
a loaded beam, is meant the algebraic sum of the moments of all the 
loads (including weight of beam) and reactions to the left or right 
of the section with respect to any point in the section. 

Rule of Signs. The rule of signs previously stated, Page 14, 
is followed; but in order to get the same sign for the bending moment 
whether computed from the right or left, the sign of the sum of the 
moments is changed when computed from the loads and reactions 
on the right. Thus for section a, Fig. 5, the algebraic sum of the 
moments of the forces, when computed from the left 



MACHINE DESIGN 


21 


= (-1,000 X 5) + (3,000 X 1) 

= — 2,000 foot-pounds 

and when computed from the right 

= (1,000 x 19) - (3,000 X 15) + (2,000 X 13) + (2,000 X 1) 

= + 2,000 foot-pounds 

The bending moment at section a is —2,000 foot-pounds. 

Again, for section b, the algebraic sum of the moments of the 
forces, when computed from the left 

= (-1,000 X 22) + (3,000 X18) - (2,000 X16) - (2,000 X 4) + (3,000 X 2) 
= — 2,000 foot-pounds 

and when computed from the right 

= 1,000 X 2 

= — 2,000 foot-pounds 

The bending moment at the section is —2,000 foot-pounds. 

It is usually convenient to compute the bending moment for a 
section from the forces to the right or left according as there are 
fewer forces (loads and reactions) on the right or left side of the 
section. 


TABLE II 

Coefficient of Elasticity (E). Moment of Inertia (I), Values of Maximum Shear (V), 

Bending Moment (M), and Defection (d). 


' l p 

Z 

W-total uniform loa-d 

• ......_) ...^ 


V-P, M*P1, d-Pl 3 + 3EL 

-- 1 -H 

V-W, M^Wl,d*Wl 3 -s-8El 

3 IP 

A 

W-total uniform load 

A " A 

A- % 

V-/ Z P, d“Pl 5 ^48EI 

V-^W, d'5Wl 3+ 384EI 

5 [-^ --*| 

6 h-H p PH 

A A 

V-P&+1, 

k k 

V»p M-Pa,d>Pa(31 z -4l)+J?4EI 

7 j r | 

8» 6 

1 W- total uniform load i 

V«XR M*I^Pl, d-Pl 3 +!92 El 

V-j4w, M&W), d-Wl 3 *384E! 
























22 


MACHINE DESIGN 


Units. It is customary to express bending moments in inch- 
pounds, but often the foot-pound unit is more convenient. To 
reduce foot-pounds to inch-pounds, multiply by twelve. 

Maximum Bending Moment. It is sometimes desirable to know 
the greatest or maximum value of the bending moment in a given case. 
This mav be obtained from Table II. 

STRENGTH OF BEAMS 

Resisting Moment of Beams. When a beam is bent the stresses 
acting over a cross-section are not the same at all points. The 
fibers on one side are stretched and therefore under tension, while 
those on the other side are shortened and under compression. In a 
simple beam the upper fibers are compressed and shortened and the 
lower fibers lengthened. In a cantilever beam the upper fibers 
are in tension, while the lower ones are compressed. This change 
in length of fibers, and therefore in the intensity of stress, proceeds 
uniformly from the bottom to the top of the beam. Since the character 
of the stress changes from tension to compression or compression to 

tension in passing from the top 
to bottom, there must be a point 
or layer in the beam which has 
neither tension nor compression 

exerted upon its fibers. This 
Fig. 8. Resistance Of Beam to Bending i ayer or surface is called the 

neutral surface and is represented in Fig. 8 by a plane NNNN. The 

central line 00 of the neutral surface is called the neutral axis and 

passes through the center of gravity of the section. The resistance 

offered to bending by a beam is B, and is equal to the bending mo¬ 
ment M, therefore 

M = S— (4) 

where S is the unit stress in the fiber at the greatest distance from the 
neutral axis; c is the distance of this fiber from the axis; and I is the 
moment of inertia, a factor determined by higher mathematics. 

Taking 8 as a safe working stress of the beam, S— should be equal to 

c 

the greatest bending moment. The ratio — is called the section 

c 












MACHINE DESIGN 


23 


*TABLE III 

Moments of Inertia, Section Moduli, and Radii of Gyration 



*In each case the axis is horizontal and passes through the center of gravity. 

modulus. Values of the moment of inertia and the section modulus 
for different cross-sections are given in Table III. 

Deflections of Beams. Sometimes it is desirable to know how 
much a given beam will deflect under a given load. In Table II 
formulas are given for the deflection in certain cases of beams and 
different kinds of loading. 

Examples. 1. A wrought-iron cantilever beam having a square 
section, 2 inches on a side, is 10 feet long. If the beam weighs 
12 pounds per foot and has a working fiber strength equal to 12,000 
pounds, what load may be safely placed on its extremity? 

Solution. Referring to Table II, the maximum bending moment 
for a uniform load is \ Wl. The weight of the beam is a uniform load 
and W = 12 X 10 = 120 pounds, / = 12 X 10 = 120 inches, then 
M x = \ Wl = JX 120 X 120 = 7,200 inch-pounds. 













































24 


MACHINE DESIGN 


The resisting moment of the beam is S —. The value of—, the sec- 

c c 

tion modulus, may be obtained from Table III and for a square 


a* 


section is—, then 


c 


a * 


2 3 8 


G 


6 G 


M = S— 

c 


8 


12,000 X — = 1G,000 inch-pounds. 


The resisting moment must oppose the moment due to the 
weight of the beam as well as the load and the resisting moment 
opposing the load is M — M x = 16,000 — 7,200 = 8,800 inch-pounds. 
From Table II the maximum bending moment due to the load is 

M = PI = P X 120 inch-pounds. 

Equating 

P X 120 = 8,800 
„ 8,800 

P = - ' ^ 2 q = 73J pounds. 

2. What is the deflection of the above beam, due only to its 
own weight? The coefficient of elasticity is 25,000,000. 

Solution. From Table II the deflection is expressed as 

d = WP + 8EI 

From Table III 

a 4 16 4 

“12 “12“ 3 


then 


d = 


120 X 120 3 
8 X 25,000,000 X $ 


= .77 inch 


STRENGTH OP SHAFTS 

Shafts. A shaft is a part of a machine or system of machines, 
and is used to transmit power by virtue of its torsional strength, or 
resistance to twisting. Shafts which are nearly always made of 
metal are usually circular in cross-section and sometimes hollow. 

Twisting Moment. Let ^4F, Fig. 9, represent a shaft with 
four pulleys on it. Suppose that D is the driving pulley and that 
B, C, and E are pulleys from which power is taken off to drive ma¬ 
chines. When the shaft is transmitting power, the portions between 








MACHINE DESIGN 


25 


the pulleys are twisted—by the twisting moment at any cross-section 
of the shaft is meant the algebraic sum of the moments of all the forces 
acting on the shaft on either side of the section, the moments being 



Fig. 9. Twisting Action on Shaft 

taken with respect to the axis of the shaft. Thus, if the forces acting 
on the shaft (at the pulleys) are P v P 2 , P 3 , and P 4 , as shown, and if 
the arms of the forces or radii of the pulleys are a v a 2 , a 3 , and a 4 , 
respectively, then the twisting moment at any section 

in BC = P t a t 

in CD = P l a l + P 2 a 2 

in DE = P x a t + P 2 a 2 — P 3 a 3 

Like bending moments, twisting moments are usually expressed 
in inch-pounds. 

Example. Let = a 2 = a 4 = 15 inches, a 3 = 30 inches, P t 
= 400 pounds, P 2 = 500 pounds, P 3 = 750 pounds, and P 4 = 600 
pounds.* What i« the value of the greatest twisting moment in the 
shaft? 

At any section between the first and second pulleys, the twisting 
moment is 

400 X 15 = 6,000 inch-pounds 
at any section between the second and third it is 

(400 X 15) -f (500 X 15) = 13,500 inch-pounds 
and at any section between the third and fourth it is 

♦These numbers were so chosen that the moment of P (driving moment) equals 
the sum of the moments of the other forces. This is always the case in a shaft rotating 
at constant speed; i. e., the power given the shaft equals the power taken off. 



26 


MACHINE DESIGN 


c 


I I 


Fig. 10. Flange 
Coupling 


(400 X 15) + (500 X 15) - (750 X 30) = - 9,000 inch-pounds 
Hence, the greatest value is 13,500 inch-pounds. 

Torsional Stress. The stresses in a twisted shaft are called 
torsional stresses. The torsional stress on a cross-section of a shaft 
is a shearing stress, as in Fig. 10, which represents a flange coupling 

in a shaft. Were it not for the bolts, one 
flange would slip over the other when either 
J part of the shaft is turned; but the bolts pre¬ 
vent the slipping. Obviously there is a ten¬ 
dency to shear the bolts off unless they are 
screwed up very tight; i. e., the material of the 
bolts is subjected to shearing stress. 

Just so, at any section of the solid shaft there is a tendency for 
one part to slip past the other, and to prevent the slipping or shearing 
of the shaft, there arise shearing stresses at all parts of the cross- 
section. The shearing stress on the cross-section of a shaft is not a 
uniform stress, its value per unit-area being zero at the center of the 
section, and increasing toward the circumference. In circular 
sections, solid or hollow, the shearing 
stress per unit-area (unit-stress) varies di¬ 
rectly as the distance from the center of 
the section, provided the elastic limit is 
not exceeded. Thus, if the shearing 
unit-stress at the circumference of a sec¬ 
tion is 1,000 pounds per square inch, 
and the diameter of the shaft is 2 inches, 
then, at J inch from the center, the unit- 
stress is 500 pounds per square inch; and 
at J inch from the center it is 250 pounds 
per square inch. In Fig. 11 the arrows 

indicate the values and the directions of the shearing stresses on very 
small portions of the cross-section of a shaft there represented. 

Resisting Moment. By resisting moment at a section of a shaft 
is meant the sum of the moments of the shearing stresses on the 
cross-section about the axis of the shaft. 

Let S s denote the value of the shearing stress per unit-area 
(unit-stress) at the outer points of a section of a shaft; d, the diameter 
of the section—if the shaft is hollow, let d equal outside diameter 



Fig. 11. Cross-Section 
of Shaft 















MACHINE DESIGN 


27 


and d x inside diameter. Then it may be shown that the resisting 
moment T is 

T = “ (5) 

where c is the distance of the most remote fiber from the neutral axis, 
and I is the polar moment of inertia. This is not the same as the 
direct moment of inertia, and must be determined by the aid of Cal¬ 
culus 

Formula for the Strength of a Shaft. As in the case of beams, 
the resisting moment equals the twisting moment at any section. 
Determining the value of I by higher mathematics, the formula 
S. I 


T = 


becomes 


for solid circular shafts, 
for hollow circular shafts, 


T = 0.1963 S a d 3 
T _ 0.1963 S s (d A - d*) 
d 


( 6 ) 

(7) 


In any portion of a shaft of constant diameter, the unit-shearing 
stress S a is greatest where the twisting moment is greatest. Hence, 
to compute the greatest unit-shearing stress in a shaft, first determine 
the value of the greatest twisting moment and substitute its value 
in the first or second equation above, as the case may be, and solve 
for S a . It is customary to express T in inch-pounds and the diameter 
in inches, S a then being in pounds per square inch. 

Examples. 1, Compute the value of the greatest shearing 
unit-stress in the portion of the shaft between the first and second 
pulleys represented in Fig. 9, assuming values of the forces and pulley 
radii as given in the example, Page 25. Suppose also that the shaft 
is solid, its diameter being 2 inches. 

Solution. The twisting moment T at any section of the portion 

between the first and second pulleys is 6,000 inch-pounds. Hence, 

substituting in the first of the two formulas above 
© 

0.1963 S a X 2 3 = 6,000 

or 

6,000 


5 .= 


= 3,820 pounds per square inch 


0.1963 X 8 

This is the value of the unit-stress at the outside portions of all 
sections between the first and second pulleys. 

2. A hollow shaft is circular in cross-section, and its outer 






28 


MACHINE DESIGN 


and inner diameters are 16 and 8 inches respectively. If the work¬ 
ing strength of the material in shear is 10,000 pounds per square inch, 
what twisting moment can the shaft safely sustain? 

Solution. The problem requires merely the substitution of the 
values of S„ d, and d x in the second of the above formulas, and solving 
for T. Thus 


T= 


0.1963 X 10,000 (16 4 -8 4 ) 
16 


= 7,537,920 inch-pounds 


PROBLEMS FOR PRACTICE 


1. Compute the greatest value of the shearing unit-stress in 
the shaft represented in Fig. 9, using the values of the forces and 
pulley radii given in the example, Page 25, the diameter of the shaft 
being 2 inches. 

Ans. 8,595 pounds per square inch. 

2. A solid shaft is circular in cross-section and is 9.6 inches 
in diameter. If the working strength of the material in shear is 
10,000 pounds per square inch, how large a twisting moment can 
the shaft safely sustain? (The area of the cross-section is practically 
the same as that of the hollow shaft of Example 2, Page 27.) 

Ans. 1,736,736 inch-pounds. 

COMBINED STRBSSES 



Formulas. Nearly all stresses may be reduced to simple tension, 
compression, shear, or torsion. In many cases complex combina¬ 
tions occur, which will not permit simple and direct application of 
the formulas for these, but it is essential to have perfect command 
of formulas 4 and 5. Assuming that the forces may be analyzed 
and the simple moment completed at the point where it is desired 
to find the strength of section, it remains only to insert the assumed 
working fiber stress of the material in the formula and to solve for 
the quantity desired. 

In cases of combined stress, the relations become more com¬ 
plicated and difficult of analysis and solution. The most common 
case is that of a shaft transmitting power, and at the same time loaded 
transversely between bearings, which is a combination of bending 
and torsion—there are very few cases of shafts in machines, which, 
at some part of their length, do not have this combined stress. In this 
case the method of procedure is to find the simple bending moment 



MACHINE DESIGN 


29 


and the simple torsional moment separately, in the ordinary way. 
Then the theory of elasticity furnishes a formula for an equivalent 
bending or an equivalent torsional moment which is supposed to 
produce the same effect upon the fibers of the material as the com¬ 
bined action of the two simple moments acting together. In other 
words, the separate moments combined in action, being impossible 
of solution in that form, are reduced to an equivalent simple moment 
and the solution then becomes the same as for the previous case.* 


K= Y+ 4 Vm 2 + T 2 


( 8 ) 


T. = M + V IP + T 2 (9) 

M , and r l\, found from these equations, are the external mo¬ 
ments, and are to be equated to the internal moments of resistance 
of the section precisely as if they were simple bending or torsional 
moments. Although either formula may be used, the latter is more 
common for shafts. 

STRENGTH OF PIPES AND CYLINDERS 

Pipes and Cylinders. Water and steam exert equal pressures 
in all directions and tend to rupture a pipe or 
cylinder longitudinally or in the direction of its 
length. The material resists this rupturing stress. 

Let p = pressure per square inch; D = 
diameter of the pipe; L = length of the pipe; t = 
thickness of the pipe; S = tensile stress. Then 
from a principle of hydraulics, the force which 
tends to cause rupture is equal to the pressure 
on a plane whose width is equal to the diameter 
of the pipe D and whose length is equal to L. 

Such a plane is indicated in Fig. 12 by abed . 

The area of the plane is LD. The force act¬ 
ing to rupture the pipe or cylinder is 

P = pLD (10) 

Formulas, The resistance of each side is equal to the area of 
metal multiplied by the stress per square inch, or tLS, and for two 
sides, 2tLS. Then, as the resistance must be at least equal to the 
force tending to rupture, 

♦The subscript “e” expresses separation from the simple moments. 



Fig. 12. Pipe Under 
Internal Pressure 




















30 


MACHINE DESIGN 


pLD = 2 tLS 
or 

pD = 2 tS 

Dividing by 2, the formula becomes 



or, since D = 2r 



and the equation may be written as 

t p thickness _ pressure per square inch 

r S radius working stress 



In the solution of problems, the above ratio will be found to be a 
convenient one. 

Example. What should be the thickness of a cast-iron pipe 
10 inches in diameter, with an internal pressure of 150 pounds per 
square inch, and 1,250 pounds per square inch as the safe working 
stress? 


Solution. 


L=JP 

r S 



5 X 150 
1250 



The pipe should, therefore, be about § of an inch thick. 


METHOD OF DESIGN 

The fundamental lines of thought and action which every de¬ 
signer follows in the solution of a problem in design are four in num¬ 
ber. The expert may carry all these in mind at the same time, 
without definite separation into a step-by-step process; but the student 
must master them in their proper sequence, and thoroughly under¬ 
stand their application. In these four fundamentals is concentrated 
the entire art of Machine Design, and when they have become so 
familiar as to be instinctively applied on any and all occasions, good 
design is the result. Experience is the only other factor which will 
facilitate still further the design of good machinery; and it cannot 
be taught, it must be acquired by actual work. 











MACHINE DESIGN 


3! 


ANALYSIS OP CONDITIONS AND FORCES 

First, take a good square look at the problem to be solved. 
Study it from all sides, view it in all lights, note the worst conditions 
which can possibly exist, the average conditions of service, and any 
special or irregular service likely to be called for. 

With these conditions in mind, make a careful analysis of all 
the forces, maximum as well as average, which may be brought into 
play. Although it is hard and sometimes impossible to determine 
exactly the forces acting on a given piece, their nature—whether 
sudden or slowly applied, rapid in action or only occurring at inter¬ 
vals—and their approximate magnitude, are always capable of 
analysis. Make a rough sketch of the piece under consideration, 
put in these forces, and go over the analysis carefully. A hasty and 
poor analysis will in the end be time lost, and, if the machine actually 
fails from this reason, heavy financial loss in material and labor will 
occur. 

Machines are nothing but a collection of parts upon which forces 
are acting directly, or parts acting as loaded beams. Where the 
force has no leverage it acts directly on the sustaining part. Forces 
acting with leverage produce a moment; the sustaining member is 
a beam, and the effect produced therein depends on the theory of 
beams. 

An example of the first is the load on a rope, the force aeting 
without leverage, and the rope, therefore, having a direct pull put 
upon it. 

An example of the second is a push of the hand on the crank 
of a grindstone. A moment is produced about the hub of the crank; 
the arm of the crank is a beam. 

THEORETICAL DESIGN 

After it is determined by careful analysis what stress the ma¬ 
chine part has to sustain, the next step is so to design it that it will 
theoretically resist the applied forces with the least expenditure of 
material. 

Machinery is often constructed with the metal of which it is 
made distributed in the worst possible manner. In places where the 
stress is heavy and a rigid member is needed, a weak, springy part 


32 


MACHINE DESIGN 


is sometimes found; while in other parts, where there are no forces 
to be resisted, or vibration to be absorbed, there seems to be a waste 
of good material. Whether in such case the analysis of the forces 
was poor, or perhaps not made at all, or whether a knowledge of 
how to design so as to resist the given forces was wholly absent, 
cannot be told. At any rate, lack of either or both is clearly shown 
in the result. 

Any member of a machine may vary in form from a solid block 
or chunk of material to an open ribbed structure. The solid chunk 
fills the requirement as far as strength is concerned, unless it is so 
heavy as to fail from its own weight. But such construction is poor 
design, except in cases where the concentration of heavy mass is 
necessary to absorb repeated blows like those of a hammer. The 
possibility of these blows should, however, have been determined 
in the analvsis; and the solid, anvil construction then becomes theo- 
retical design for that analysis. 

For steadily applied loads an open, ribbed, or hollow box struc¬ 
ture can be made which will distribute the metal where it is theoretically 
needed, and each fiber will then sustain its proper share of the load. 
In this way weight, cost, and appearance are heeded; and the service 
of the piece is as good as, and probably better than, it would be with 
the clumsy, solid form. 

There is no such thing as putting too much theory into the 
design of machinery. The strongest trait which an engineer can 
have is absolute faith in his analysis and calculations, and their 
reproduction in his theoretical design. Theoretical design is an 
indication of scientific advance in the art, and some of the greatest 
steps of progress which have been made in recent years have been 
accomplished through a purely theoretical study of machine structure. 

PRACTICAL MODIFICATION 

It will never do to be satisfied with theoretical design when 
it is not in accord with modern commercial and manufacturing con¬ 
siderations. Hence the next step after the determination of the 
theoretical design is the study of it from the producing standpoint. 

All theoretical design viewed from the business standpoint is 
worthless, unless it has been subjected to the test of cheap and 
efficient production. Each machine detail, though correct in theory, 


MACHINE DESIGN 


33 


may yet be improperly shaped and unfit for the part it is to play in the 
general scheme of manufacture. 

The conditions here involved are changeable. What is good 
design in this decade may be bad in the next. In this light the 
designer must be a close student of the signs of the times; he must 
follow the march of progress, closely applying existing resources, 
conditions, and facilities, otherwise he cannot produce up-to-date 
designs. The introduction of new raw materials, the cheapening of 
production of others, the changing of shop methods, the use of special 
machinery, the opening of new markets, the development of new 
motive agents—all these and many others are constantly demanding 
some modification in design to meet competition. 

Illustrative of this, note the change which has been wrought 
by the development of electric power, the rise and decline of the 
bicycle business, the present manufacture of automobiles, the last 
named especially with reference to the development of the small 
motive unit, the gasoline engine, the steam engine, etc. The design 
of much machinery has been materially changed to meet the exact¬ 
ing demands of these new enterprises. 

Practical modifications of design necessary to meet the limi¬ 
tations of construction in the pattern shop, foundry, and machine 
shop are of daily application in the designer’s work. He must keep* 
in his mind’s eye at all times the workmen and the processes they 
use to create his designs in metal in the shop. 

“How can this be made?” “Can it be made at all?” “Can 
it be made cheaply?” “Will it be simple in operation after it is 
made?” “Can it be readily removed for repair?” “Can it be lub- 
* ricated?” “How can it be put in place?” “How can it be gotten 
out?” “Will it be made in small quantities or large?” “Will it sell 
as a special or standard machine?” etc., etc. 

The consideration of such questions as these is a practical 
necessity from a business standpoint, as no other factor affects the 
design of machinery more. Designs which cannot be built as business 
propositions are no designs at all. 

The student, it is true, may not have the extended shop knowl¬ 
edge which is essential to this; but he can do much for himself by 
visiting shops whenever possible, getting hold of shop ways of doing 
things, and invariably treating his work as a business matter. Though 


34 


MACHINE DESIGN 


a man may not be a pattern maker, moulder, blacksmith, or machinist, 
yet he can soon gain ideas of the processes of each of these branches 
which will be of immense advantage to him in his designing work. 

DELINEATION AND SPECIFICATION 

Delineation and specification mean the clear and concise repre¬ 
sentation of the design by mechanical drawings, and are as much a 
part of the routine method of Machine Design as the three preceding 
fundamentals. The mere act of putting the results of mechanical 
thinking on paper is one of the greatest helps for bringing the think¬ 
ing machinery into systematic and definite action. A designer never 
thinks very long without drawing something, and the student must 
bring himself to feel that a drawing in its first sense is a means of 
helping his own thought, and must freely use it as such. 

In its second and final sense, the drawing is an order and speci¬ 
fication sheet from the designer to the workman. Design which 
stops short of exact, finished delineation in the form of working 
shop drawings, is only half done. In fact, the possibility of a piece 
being thus exactly drawn is often the crucial test of its feasibility 
as a part of a machine. It is easy to make general outlines, but it 
is not so easy to get down to finished detail. It is safe to say that 
there is no one thing productive of more trouble, delay, and embar¬ 
rassment, and waste of time and money in the shop, when there need 
be none from this cause, than a poor detail drawing. The efficiency 
of the process of design is not fully realized, and failures are often 
recorded where there should be success, merely because the in¬ 
definiteness permitted by the designer in the drawings naturally 
transmitted itself to the workman, and he in turn produced a part 
indefinite in form and operation. 

The actual process of drawing in the development of a design 
may be outlined as follows: 

Rough sketches merely representing ideas, not drawn to scale, are first 
made. These are of use only so far as the choice of mechanical ideas is con¬ 
cerned, and to carry preliminary dimefisions. 

Following these sketches, comes a layout to scale, of the favored sketch, 
a working out of the relative sizes and location of the parts. This drawing 
may be of a sketchy nature, carrying a principal dimension here and there 
to fix and control the detailed design. In this drawing the design is developed 
and general detail worked out. The minute detail of the individual parts 
is, however, left to the subsequent "working drawing. 


MACHINE DESIGN 


35 


This layout drawing may now be turned over to an expert draftsman, 
or detail designer, who picks out each part, makes an exact drawing of it, 
studying every little detail of its shape, and finally adds complete dimensions 
and specifications so that the workman is positively informed as to every point 
of its construction. 

Detail Drawings. Drawings of individual parts and sections 
showing the internal construction and sectional shape are sometimes 
necessary. These are called detail drawings. 

Sometimes it is necessary to show sections in order that the inter¬ 
nal construction or sectional shape may be easily understood. These 
sections are usually drawn through the axis, or center, but it is some¬ 
times advisable to show sections of other portions. Where the 
drawing shows a section, the portions of metal or wood supposed 
to be cut are covered with parallel lines at equal distances and usually 



CAST IRON 


WROUGHT IRON 




LEAD OR BA8BIT 




VULCANITE 


WOOD, 


Fig. 13. Conventional Cross-Hatching 



BRICK 


oblique. These sections are called hatched, cross-hatched or 
simply sectioned. The character of the lines—full, dotted, broken, 
light, or heavy—indicate the material supposed to be cut. One kind 
indicates cast iron, another steel, another brass, etc. There is no 
standard for cross-hatching, different draftsmen using lines of various 
character. There is likely to be a confusion unless the parts have 
the name of the material printed on or near it, or a key is provided. 

Fig. 13 shows the lines as generally used, those representing 
cast iron, brass, wood, and lead being almost universal; the others 
are subject to more change. The lines may run from left to right, 
or right to left; in case two or more parts of the same metal are brought 























36 


MACHINE DESIGN 


together it is necessary to avoid confusion by varying the direction 
and angles of the lines. If the cross-hatching were to be the same, 
the parting line would be confused and one might think it all one 
piece. 

When drawing designs of the details, it is well to make them 
as large as convenient. The scales in general use are full size or 
half size; 3 inches or 1^ inches = 1 foot. A drawing is seldom 
made by such scales as 2 inches or 4 inches = 1 foot. 

A working drawing is one that shows all the dimensions of 
an object in such a manner that it may be made by reference to 
the drawing. Usually three views are sufficient—two elevations, 
taken at right angles to each other, and a horizontal projection or 
plan. Often the plan may be dispensed with if the object repre¬ 
sented is of simple form. Sections to show the interior construction 
are sometimes added. 

In this discussion, the dimensions of some of the drawings are 
given in letters instead of figures, so that the relation between the 
different dimensions may be explained. In general, however, this 
should not be done. The dimensions placed upon a drawing should 
be in figures, for letters without an explanatory table convey no idea 
of the size. 

General Drawing. The last step in the process of design of a 
machine is the making of the assembled or general drawing. This 
is built up piece by piece from the detail drawings, and serves 
as a last check on the parts which are to be put together. This 
drawing may be a cross-section or an outside view. In any case it 
is not wise to try to show too much of the inside construction by 
dotted lines, for if this be attempted, the drawing soon loses its 
character of clearness, and becomes practically useless. A general 
drawing should clearly hint at, but not specify, detailed design. It is 
just as valuable a part of the design as the detail drawing, but it 
cannot be made to answer for both with any degree of success. A 
good general drawing has plenty of view^s and an abundance of 
cross-sections, but few dotted lines. 

The functions which the general drawing may serve are many 
and varied. Its principal usefulness is, perhaps, in showing to 
the workman how the various parts go together, enabling him to 
sort out readily the finished detail parts and assemble them, finally 


MACHINE DESIGN 


37 


producing the complete structure. Otherwise the making of a 
machine, even with the parts all at hand, would be like the putting 
together of the many parts of an intricate puzzle, and much time 
would be wasted in trying to make the several parts fit, with per¬ 
haps never complete success in giving each its absolutely correct 
location. 

The general drawing also gives valuable information as to the 
total space occupied by the completed machine, enabling its loca¬ 
tion in a crowded manufacturing plant to be planned for, its con¬ 
nection to the main driving element arranged, and its convenience 
of operation studied. 

In some classes of work it is a convenient practice to. letter 
each part on the general drawing, and to note the same letters on 
the specification or order sheet, thus enabling the whole machine 
to be ordered from the general drawings. This, although a very 
excellent method in certain lines of work, makes the general draw¬ 
ing quite useless in others. 

Merely as a basis for judgment of design, the general drawing 
fulfills an important function in any class of work, for it approaches 
the nearest possible to the actual appearance that the machine will 
have when finished. A good general drawing is, for critical pur¬ 
poses, of as much value to the expert eye of the mechanical engineer 
as the elaborate and colored sketch of the architect is to the house 
builder or landscape designer. 

From the above it is readily understood that the general draw¬ 
ing, although a mere putting together of parts in illustration, is yet 
of great assistance in producing finished and exact Machine Design. 

LUBRICATION 

Friction. The parts of a machine which have no relative 
motion with regard to each other are not dependent upon lubrication 
of their surfaces for the proper performance of their functions. In 
cases where relative motion does occur—as between a planer bed and 
its ways, a shaft and its bearing, or a driving screw and its nut— 
friction, and consequent resistance to motion, will inevitably occur. 
Heat will be generated, and cutting or scoring of the surfaces will 
take place, if the surfaces are allowed to run together dry. 


38 


MACHINE DESIGN 


This difficulty, which exists with all materials, cannot be over¬ 
come, for it is a result of roughness of surface, characteristic of the 
material even when highly finished. The problem of the designer, 
then, is to take conditions as he finds them, and, as he cannot change 
the physical characteristics of materials, so choose those which are 
to rub together in the operation of the machine that friction will 
be reduced to the lowest possible limit. Now it fortunately happens 
that there are certain agents like oil and graphite, which seem to fill 
up the hollows in the surface of a solid material, and which themselves 
have very little friction on other substances. Hence, if a machine 
permits by its design an automatic supply of these lubricating agents 
to all surfaces having motion between them, friction may be reduced 
to the lowest limit. 

If this full supply of lubricant be secured, and the parts still 
heat and cut, then the fault may be traced to other causes, such as 
springy surfaces, localization of pressure, or insufficient radiating 
surface to carry away the heat of friction as fast as it is generated. 

Lubricant. Lubricating agents are of a nature running from 
the solid graphite form to a thick grease, then to a heavy dark oil, 
and finally to a thin, fluid oil flowing as freely as water. The solid 
and heavy lubricants are applicable to heavily loaded places where 
the pressure would squeeze out the lighter oils. Grease, forced 
between the surfaces by compression grease cups, is an admirable 
lubricator for heavy machinery under severe service. High-speed 
and accurate machinery, lightly loaded, requires a thin oil, as the 
fits would not allow room for the heavier lubricants to find their 
way to the desired spot. The ideal condition in any case is to have 
a film of lubricant always between the surfaces in contact, and it is 
this condition at which the designer is always aiming in his lubricating 
devices. 

Means for Lubrication. Oil ways and channels should be direct, 
ample in size, readily accessible for cleaning, and distributing the 
oil by natural flow over the full extent of the surface. Hidden and 
remote bearings must be reached by pipes, the mouths of which 
should be clearly indicated and accessible to the operator of the ma¬ 
chine. Such pipes must be straight, if possible, and readily cleaned. 

There is one practical principle affecting the design of methods 
of lubrication of a machine which should be borne in mind. This 


MACHINE DESIGN 


39 


is, “Neglect and carelessness by the operator must be provided for/’ 
It is of no use to say that the ruination of a surface or hidden bear¬ 
ing is due to neglect by the operator, if the means for such lubrica¬ 
tion are not perfectly obvious. This is “locking the door after the 
horse is stolen.” The designer has not done his duty until he has 
made the scheme of lubrication so plain that every part must receive 
its proper supply of oil, except by gross and willful negligence for 
which there can be no possible just excuse. 

Materials Employed in Construction. When stress is induced 
in a piece, the strain is practically proportional to the stress for all 
values of the stress below the elastic limit of the material; and when 
the external load is removed the strain will entirely disappear, or 
the recovering power of the material will restore the piece to the 
original length. 

Now it is found that if a piece is to last in service for a long 
time without danger of breakage, it must not be permitted to be 
stressed anywhere near the elastic limit value. If it is, although 
it will probably not break at once, it is in a dangerous condition, and 
not well suited to its requirements as a machine member. The 
technical name for this weakening effect is “ fatigue .V It is further 
found that the fatigue due to this repeated stress is reached at a lower 
limit when the stress is alternating in character than when it is not. 
In other words, if first a pull and then a push is exerted upon a piece 
it will be first in tension and then in compression; this alternation of 
stress repeated too near the elastic limit of the material will fatigue it, 
or wear out the fibers, and it will finally fail. If, however, a pull is 
first exerted on the piece with the same force as before, and then let 
go, the piece will be in tension and then entirely relieved; such repeti¬ 
tion of stress will finally fatigue the material, but not so quickly as in 
the first case. Experiments indicate that it may take twice as many 
applications in the latter case as in the former. 

The working stress of materials permissible in machines is 
based on the above facts. The breaking strength divided by a liberal 
factor of safety will not necessarily give a desirable working stress. 
The question to be answered is, “Will the assumed working fiber 
stress permit an indefinite number of applications of the load without 
fatiguing the material?” 

Hence it is seen that the same material may be safely used under 


40 


MACHINE DESIGN 


different assumptions of working stress. For example, a rotating 
shaft, heavily loaded between bearings, acts as a beam which in 
each revolution is having its particles subjected, first to a maxi¬ 
mum tensile stress, and then to a maximum compressive stress. 
This is obviously a very different stress from that which the same 
piece would receive if it were a pin in a bridge truss. In the former 
is a case where the stress on each particle reverses at each revolu¬ 
tion, while in the latter the same stress merely recurs at intervals, 
but never becomes of the opposite character. For ordinary steel, a 
value of 8,000 would be reasonable in the former case, while in the 
latter it may be much higher with safety, perhaps nearly double. 

From the facts stated above, it is evident that exact values for 
working fiber stress cannot be assumed with certainty and applied 
broadly in all cases. If the elastic limit of the material is definitely 
known, a working value can be based on that. Data on the strength of 
materials is available in any of the handbooks, and should be con¬ 
sulted freely by the student; it will be found somewhat conflicting, 
but will assist the judgment in coming to a conclusion. 

The principal materials used in the construction of machinery 
are cast and wrought iron, steel, copper, wood, brass, and other 
alloys. 

Cast Iron. Cast iron is used to a considerable extent in the con¬ 
struction of machines. For the heavy, massive parts—the frames of 
lathes, steam-engines, planers, etc., for example—it is the best ma¬ 
terial. It is not suitable for parts requiring strength, elasticity, or 
those subjected to shocks. For this reason piston-rods, connecting- 
rods, shafts, etc., are usually made of steel or wrought iron. 

Many complicated shapes that cannot be forged are readily 
cast. The ease with which parts may be given the desired shape 
makes cast iron valuable. 

Cast iron contains 3 to 4J per cent of carbon with a little silicon. 
The hard and white varieties are used in the manufacture of wrought 
iron. The gray irons are used in the foundry. 

Cast iron is made into the desired forms by melting it in a cupola 
and pouring into moulds. The moulds are made in sand or loam 
from patterns of pine wood. Patterns are made a little larger than 
the required casting because iron in solidifying contracts about J 
inch per foot in each direction. This contraction is called shrink - 


MACHINE DESIGN 


41 


age. In making a pattern a shrinkage rule is used which is about 
i inch longer per foot than the standard. 

Castings are likely to be put into a state of internal stress be¬ 
cause of contraction when cooling. If some parts of the casting 
contract more than others, the casting may become twisted. Thin 
parts of the castings solidify first. The contraction of the fluid 
parts strains the portions already set and their resistance to deforma¬ 
tion causes stresses to be set up in the parts which are solidifying. 

For example, the form shown in Fig. 14 has a rigid flange sur¬ 
rounding the inner part. If the contraction of the cross-piece takes 
place more slowly than the rim, it is likely to fracture. In a thick 
cylinder, as shown in section in Fig. 15, the outer portions solidify 
and begin the contraction. The contraction of the inner induces 




Fig. 14. Casting with 
Flange 



Fig. 15. Cylindrical 
Casting 


pressure in the outer portion, which being rigid causes a resistance 
to contraction of the inner layers and puts them in tension. A cylin¬ 
der so constructed is not strong to resist bursting pressure. If the 
cylinder is cast while water circulates through the core, the reverse 
distribution of initial strains is set up. This insures a stronger 
cylinder because the inner layers are in a state of compression and 
the outer portions are in tension. 

The arms of pulleys may be broken by tension if the rim is thin 
and rigid. If the arms set first the rim may break near them. To 
have successful castings, the designer must carefully consider the 
dimensions of the various parts. 

On account of these initial strains, that cannot be calculated, 
cast iron is unreliable. Cast-iron structures usually have excessive 
dimensions to insure safety. 























42 


MACHINE DESIGN 


Chilled castings are cooled rapidly during solidification, thus 
preventing the graphite from separating from the iron. This causes 
the iron to become harder. In order to chill the cast iron, the mould 
is made of, or lined with, this same material. The mould which is 
lined with loam for protection, is a good conductor of heat and the 
molten cast iron is cooled or chilled during solidification. The 
chilling usually extends to a depth of J to f of an inch from the sur¬ 
face, the interior remaining soft. 

Malleable cast iron is made by surrounding castings with oxide 
of iron, powdered red hematite, or peroxide of manganese, keeping 
them at a high temperature for a considerable time according to the 
size of the casting. The elimination of carbon converts the cast 
iron into a crude form of wrought iron. Malleable castings will 
stand blows better than ordinary castings. 

Cast iron is stronger than wrought iron when under pressure; 
but it is much weaker under tension and impact. It is such an un¬ 
certain metal on account of its variable structure that stresses are 
always kept low, say, from 3,000 to 4,000 for nonreversing stress, 
and 1,500 to 2,500 for reversing stress. 

Wrought Iron. Wrought iron is made from cast iron by eliminat¬ 
ing part of the carbon. It is strong and tough and can easily be welded. 
For these reasons it is used for parts of machines and structures re¬ 
quiring strength and of simple form. Wrought iron parts are 
shaped by forging and finished in the machine shop, steam hammers 
being used on the heavy portions. 

Wrought iron is rolled into plates, round and square bars, 
angle, tee, channel, I-beam sections, etc. Large wrought iron 
structures are built up of bars or plates riveted or bolted together. 

Wrought iron that has been rolled when cold has a greater 
tensile strength than before rolling; but its ductility and toughness 
is reduced. Annealing, or heating the iron to a red heat and allow¬ 
ing it to cool slowly, restores it to the original condition. 

Compression of iron when cold increases its strength but reduces 
its ductility and toughness; annealing reduces strength and increases 
toughness and ductility. If the iron is rolled or hammered when hot, 
compression and annealing are carried on at the same time. Average 
wrought iron may be used for a load of from 8,000 to 10,000 lbs. per 
sq. in., nonreversing, and from 6,000 to 7,000 lbs. per sq. in., reversing. 


MACHINE DESIGN 


43 


Steel. Steel is by far the most useful material used in machines. 
It is not found in nature but is an alloy or mixture of iron, carbon, 
silicon, manganese, phosphorus, nickel, tungsten, etc. The strength, 
ductility, and other characteristics of steel depend upon the propor¬ 
tion of iron, carbon, and other materials in the mixture. 

It is now successfully cast by the use of silicon, aluminum, and 
other elements and internal stresses are reduced by prolonged an¬ 
nealing. It can be welded, but greater care is necessary than in the 
welding of wrought iron. 

Tempering greatly increases the usefulness of steel, since it 
becomes hard if heated and cooled suddenly. With good steel 
almost any desired hardness may be obtained. The steel is heated 
to the temperature indicated by the color of the oxide which forms 
at its surface and is then quenched in oil or water. Hardness makes 
it suitable for cutting tools. When tempered it is hard, strong, has 
high elastic limit and little ductility. 

Alloy steels are now extensively used for special purposes. The 
chief alloy steels are manganese steel, tungsten steel, nickel steel, 
chrome steel, molybdenum steel, and vanadium steel. 

Manganese steel is very ductile and possesses great hardness, 
but is too hard to be shaped by cutting tools. It is extensively used 
in rock crushing machinery. 

Tungsten steel holds its temper very well under heat and retains 
magnetism. 

Nickel steel has great hardness, strength, and ductility. It is 
much used for shafting. 

Chrome steel when properly treated has a high elastic limit and 
great hardness. It is used in dies and in stamp mill machinery. 

Molybdenum steel is very similar in its characteristics to tungsten 
steel. 

Vanadium steel possesses endurance under shocks. Its strength 
under steady loads is about the same as that of carbon steel. At 
present vanadium steel is extensively used for automobiles. 

With but a general knowledge of the elastic limit, ordinary 
steel is good for from 12,000 to 15,000 pounds per square inch non¬ 
reversing stress, and 8,000 to 10,000 reversing stress. 

Copper. Copper is a reddish metal of great ductility and mal¬ 
leability. It is usually rolled or hammered into shape because it 


44 


MACHINE DESIGN 


does not cast well. Copper can be welded, but as it requires con¬ 
siderable care to make a good joint, pieces are more often joined by 
brazing. It can be drawn into wire. The tensile strength of cast 
copper is about 20,000 pounds per square inch; of forged copper 
about 30,000 pounds per square inch. 

Hammering, rolling, and wire-drawing increases the tensile 
strength, but makes it hard and brittle. It can be made soft and 
tough by annealing. It is expensive and is used for wire, fittings, 
and tubing. Its strength is less than that of wrought iron and de¬ 
creases rapidly with rise of temperature. 

Aluminum. Aluminum is a soft, ductile, malleable metal of 
bluish white color. It is very light; next to magnesium the lightest 
of the useful metals. Its strength is about one-third that of wrought 
iron. Aluminum casts well, the shrinkage being about the same 
as brass. The readiness with which aluminum unites with other 
metals makes it valuable for alloys. It can be electrically welded 
but does not solder well. 

Bronze. Bronze, or gun-metal, is an alloy of copper and tin— 
about 90 parts copper and 10 parts tin. It makes good castings. 
Bronze is harder and less malleable than copper. It is used for 
bearings because it is softer and wears faster than wrought iron 
or steel shafts. 

The hardness of bronze depends upon the proportion of tin; 
to increase hardness increase the amount of tin. An alloy of 92 
parts copper and 3 parts tin is a soft bronze used for gear wheels. 

Phosphor-bronze is made by mixing 2 or 3 per cent of phos¬ 
phorus with ordinary bronze. 

Manganese bronze, called white bronze, is an alloy of ordinary 
bronze and ferro-manganese. Like phosphor-bronze it is used in 
marine work, because it resists the corroding action of sea-water. 
Manganese bronze is equal in tensile strength and toughness to mild 
steel and can be easily forged. 

Brass. The alloy of copper and zinc is called brass; sometimes 
tin and a little lead are added. For bearings it has about 60 per cent 
copper, 10 per cent zinc, and 30 per cent tin and lead. Naval 
brass has 62 per cent copper, 1 per cent tin and 37 per cent zinc. 
Red brass consists of about 37 per cent copper and for the rest 
about equal parts of tin, zinc, and lead. Brass is used for bearings, 


MACHINE DESIGN 


45 


wire, fittings, and ornamental work. Its tensile strength is about 
23,000 pounds per square inch. 

Fusible Alloys. Fusible alloys are made of tin, lead, and bismuth. 
The melting point varies with the percentages of the various con¬ 
stituents. If made of 2 parts lead and 1 part tin, it melts at 475° F.; if 
1 part lead, 1 part tin, and 4 parts bismuth, the melting point is about 
200° F. An alloy of 1 part cadmium, 4 parts bismuth, 1 part tin, 
and 2 parts lead melts at 165° F. 

Bearing Alloys. The principal constituents of bearing alloys are 
copper, tin, lead, zinc, antimony, and aluminum. The bronzes 
contain a large per cent of copper. A good bearing alloy is made 
of copper, 77 parts by weight, tin 3 parts, and lead 15 parts. 

Babbitt metals have various proportions; hard babbitt having 
about 89 per cent tin, 4 per cent copper, and 7 per cent antimony. 

There are many other alloys containing the metals in varying 
proportions according to the intended use. 

Wood. Wood is but little used in machine construction. Soft 
woods like pine are used for patterns; hard varieties, oak and lignum- 
vitae for examples, are used for bearings. Sometimes levers are 
made of wood and the pulleys of some lathes are constructed of the 
same material. The cogs of mortise wheels are often made of beech 
or horn-beam. 

With the foregoing information as a guide and the special con¬ 
ditions controlling each case carefully studied, reasonable limits 
may be assigned for working stresses of the various materials used 
in machines. 


BOLTS, STUDS, NUTS, AND SCREWS 

NOTATION—The following notation is used throughout the chapter on Bolts, 

Studs, Nuts, and Screws: 


a = Area at root of thread (sq.in.) 
d = Diameter of bolt (inches) 
d, = Diameter at root of thread (in¬ 
ches) 

e = Efficiency of screw and nut 
H = Height of nut (inches) 

I = Initial axial tension (lbs.) 
k — Allowable bearing pressure on 
surface of thread (lbs.per sq.in.) 

L = Lead, or distance nut advances 
along axis in one revolution 
(inches) 


l = Length of wrench handle (inches) 

n = Number of threads in nut = — 

P 

P = Axial load (lbs.) 
p = Pitch of thread, or distance be¬ 
tween similar points on adja¬ 
cent threads, measured paral¬ 
lel to axis (inches) 

S — Fiber stress (lbs. per sq. in.) 

W = Load on bolt(lbs.) 


46 


MACHINE DESIGN 


Analysis. A bolt is simply a cylindrical bar of metal upset 
at one end to form a head, and having a thread at the other end, 



Fig. 16. A stud is a bolt in which the head is replaced by a thread; 
or it is a cylindrical bar threaded at both ends, usually having a small 
plain portion in the middle, Fig. 17. The object of bolts and studs 
is to clamp machine parts together, and yet permit these same parts 
to be readily disconnected. The bolt passes through the pieces to 
be connected, and, when (ightened, causes surface compression be¬ 
tween the parts, while the reactions on the head and nut produce 



nected parts and are screwed into the other, the stud remaining in 
position when the parts are disconnected. 

As all materials are elastic within certain limits, the action of 
a bolt in clamping two machine parts together, more especially if 
there is an elastic packing between them, may be represented 











































































MACHINE DESIGN 


47 



diagrammatically by Fig. 18, in which a spring has been introduced 
to take the compression due to screwing up the nut. Evidently 
the tension in the bolt is equal to the force necessary to compress 
the spring. Now, suppose that two weights, each equal to ^ IF, are 
placed symmetrically on either side of the bolt, then the tension in the 
bolt will be increased by the added weights if the bolt is perfectly rigid. 
The bolt, however, stretches; hence some 
of the compression on the spring is relieved 
and the total tension in the bolt is less than 
IF + 7, by an amount depending on the 
relative elasticity of the bolt and spring. 

Suppose that the stud in Fig. 17 is one 
of the studs connecting the cover to the 
cylinder of a steam engine, and that the 
studs have a small initial tension; then 
the pressure of the steam loads each stud, 
and if the studs stretch enough to relieve 
the initial pressure between the two surfaces, 
then their stress is due to the steam pressure 
only; or, from Fig. 18, when I = W; the 
initial pressure due to the elasticity of the 

joint is entirely relieved by the assumed stretch of the studs. 
Except to prevent leakage, it is seldom necessary to consider the 
initial tension, for the stretch of the bolt may be counted on to 
relieve this force, and the working tension on the bolt is simply the 
load applied. 

For shocks or blows, as in the case of the bolts found on the 
marine type of connecting-rod end, the stretch of the bolts acts like 
a spring to reduce the resulting tensions. So important is this feature 
that the body of the bolt is frequently turned down to the diameter 
of the bottom of the thread, thus uniformly distributing the stretch 
through the full length of the bolt, instead of localizing it at the 
threaded parts. 

In tightening up a bolt, the friction at the surface of the thread 
produces a twisting moment, which increases the stress in the bolts, 
just as in the case of shafting under combined tension and torsion; 
but the increase is small in amount, and may readily be taken care 
of by permitting low values only for the fiber stress. 


Fig. 18 . Action of 
Elastic Packing 




















48 


MACHINE DESIGN 


In a flange coupling, bolts are acted upon by forces perpen¬ 
dicular to the axis, and hence are under pure shearing stress. If 
the torque on the shaft becomes too great, failure will occur by the 
bolts shearing off at the joint of the coupling. 

A bolt under tension communicates its load to the nut through 
the locking of the threads together. If the nut is thin, and the num¬ 
ber of threads to take the load few, the threads may break or shear 
off at the root. With a V-thread there is produced a component 
force, perpendicular to the axis of the bolt, which tends to split 
the nut. 

In screws for continuous transmission of motion and power, 
the thread may be compared to a rough inclined plane, on which a 
small block, the nut, is being pushed upward by a force parallel to 
the base of the plane. The angle at the bottom of the plane is the 
angle of the helix, or an angle whose tangent is the lead divided 
by the circumference of the screw. The horizontal force corre¬ 
sponds to the tangential force on the screw. The friction at the 
surface of the thread produces a twisting moment about the axis of 
the screw, which, combined with the axial load, subjects the screw 
to combined tension and torsion. Screws with square threads are 
generally used for this service, the sides of the thread exerting no 
bursting pressure on the nut. The proportions of screw thread 
for transmission of power depend more on the bearing pressure than 
on strength. If the bearing surface be too small and lubrication 
poor, the screw will cut and wear rapidly. 

Theory. A direct tensile stress is induced in a bolt when it 
carries a load exerted along its axis. This load must be taken by 
the section of the bolt at the bottom of the thread. If the area at 


the root of the thread is 



and if S is the allowable stress per 


square inch, then the internal resistance of the bolt is 
ting the external load to the internal strength 


Srrd* 


Equa- 


W = Sa 


Sxd, 2 

~4 — 



For bolts which are used to clamp two machine parts together 
so that they will not separate under the action of an applied load, 
the initial tension of the bolt must be at least equal to the applied 





MACHINE DESIGN 


49 


load. If the applied load is W, then the parts are just about to 
separate when I = W. Therefore, the above relation for strength 
is applicable. As the initial tension to prevent separation should 
be a little greater than W, a value of S should be chosen so that 
there will be a margin of safety. For ordinary wrought iron and 
steel, *S may be taken at 6,000 to 8,000. 

If, however, the joints must be such that there is no leakage 
between the surfaces, as in the case of a steam cylinder head, and 
supposing that elastic packings are placed in the joints, then a much 
larger margin should be made, for the maximum load which may 
come on the bolt is I + W, where W is the proportional share of 
the internal pressure carried by the bolt. In such cases S = 3,000 
to 5,000, using the lower value for bolts of less than f-inch diameter. 



Fig. 19. Standard Bolt, Nut, and Thread 


Table IV will be found very useful in proportioning bolts with 
U. S. standard thread for any desired fiber stress. Standard dimen¬ 
sions of bolt, nut, and thread are given in Fig. 19. 

To find the initial tension due to screwing up the nut, the length 
of the handle of an ordinary wrench, measured from the center of 
the bolt, is assumed as about 16 times the diameter of the bolt. 
For one turn of the wrench a force F at the handle would pass over 
a distance 2x1, and the work done is equal to the product of the 
force and space, or F X 2xl. At the same time the axial load P 
would be moved a distance p along the axis. Assuming that there 













































TABLE IV 

Strength of Bolts— U. S. Standard Thread 


50 


MACHINE DESIGN 

i 


Approximate Shearing Strength 
Full Bolt Diameter 
(In Hundreds of Pounds) 

•hi -bs J9d 
•sqi 000'i 
ay 

«t»MO 

•xfCOC-O 

»H*-HfHC > 'IM , «$<i{5C'*a0©eNlTj<©O5C\Jt— «H*-<©aOC'*l'»OOOi»-*'1‘t-CCt» 
i-*iH»-<»~ti-<«')CN»eO»*^»<lO©C'*00©rHC')CiOCO© 

T* rH 

•hi ’bs J9d 
•sqi 000‘9 
ay 

aoNMo 

0)ffl®00 

e»-<*coa>e'i»«ooco<ot-©'*o><©T}<'«*c£i«ooo« < t<«o-<j*cot-oo-'i<i-f'i«e»ooi««© 

i-*i-ii-HC'JcOTjt©c-oo©e'i-"*©ooc'0©irt«'Jost~coio»«»n©t-cv>© 

tHr-ti~<r-li-<S'10JCiO'«*-«1'»OCOt'-00©©i-<'«J<© 

HHH 

•m -bs J9d 
•sqi 000‘S 

ay 

lO CO <M -H iH O 

•^oovniooo© 

C*COl£ , 5t>©'>f5C'3©©©*M'»*00'*t*©COt'-©»(5t-CClC5i-tCSIOO©»f>©«'JOO'<»< 

lJ^HCMMCi5lO<Ot-0O©«^MiO©'^a>»O^H00i«C^©OT0OCOCO^ 

fH 

*ui *bs J9d 
•sqi 000 

av 

© Tt< © © -^ © 

© © rt< © 00 © © 

»4fO'«J<©t^05CvioO'^<»H©©©^M©©©©©t»OOSM>rt«MCV5t'©©\0©© 

r -<iH«NlCOTt''**l«t'«00©i-ICvi»ft©COOOcOOO'»J<©©eO©aO©e'J 

i-liHr-lr-l«')C')COCO'»»<UOir5©t'»t-©*-< 

Approximate Tensile Strength 
at Bottom of Thread 
(In Hundreds of Pounds) 

•m ‘bs J9d 
•sqi 000‘01 
ay 

O ^ 00 CO o 
(D»Ol>fOO 

^i^doi^ooowLOO)oo'vDaM^MONOOTj<o^oooocio0Tf^ 
pHrHCMCO^lOCOt-OCMlClr-OCOrHt—CO^COOCOC5fOCD(rat-C^O 
^^^^CMC^COCO^i^^t-OOOirHCMTf^OJtfO 

HHHr^H 

•tn ’bs J9d 
•sqi 000 ‘L 

av 

co oo m co o 

OC rH t- m tT- o 

rHC0^CD06f-H^rH0500 00iA^Oa>e0C^^00C5CMrHC^00CQC0fH00f0C0t^b» 
iHiHCSCMCO^lOt^CTaOC^^cO^H^C^GOCDCQOaiOiOOaSO^fH 

*H *4 t* 

•m ‘bs J9d 
•sqi 000‘9 

ay 

© © t- © ©os © 

©t-©©oco© 

i~'ca'f’iQr-oic'iooiac«^t-'<»«t-e'»coc'joot~« > 3COco©e>j©-<><oo©i-icO'«i<'«* 

rHtHCMCC'l">l'©t«©©CMC000CMt-eM©lrti-i©C'©lO-^<K500 

rHi^i-li-HCNIC'JCOCO-"*lC5»O©t-OO0Ji^eO 

•m -bs J9d 
•sqi 000‘S 

ay 

Ot-OJftCOO 

MMCOCOC^HO 

▼Hcico^txiooo^'^tr-^cifo^oooo^iotoioocQotoov^kft'^ajoocMai 

TH^HCaC4COOOlOtPt-OOOpHlOOOCOt>-fOt-Cr?a5C^eOOOOO^ 

’HI • bs J9d 

•sqi 000‘t 
ay 

ao«'J^co©00-'»'© 

©oot-c-©'*©© 

^H^(Mco©©oocMt-o»oo^HCMiHi-(©'^c^moo'»foO'<»<e , <«'^«©iMt-t-©©fo 
r*i-HC'aC<lCOTl<»r5C0t-00©C'l'*J<00.H©©'*J<©©©CDcOt— CM 

rH.-i.-ic<ievicoeocO'*»OiCi©t-© 

Areas 

pB9jqx 
jo taowog 

co-^t-©M©©©«^m©©m©T^'»i<io©s s 5< s J(M«c5i-iur5'^<©f(9Tj<e , i>cot-© 

©©©©.-<i-l«')C0T»<lCS©00©e>«inr-©C0©C-©-«J<l£5O©CJr0t'C'ac~C>q© 

»Hi-C^r*C , »C'4COCO'l<iOCOC»00©^H<M-^ima5 C Y5 

HHHHHN 

iCpog 

aioa 

wao*-<ici©»OFH'«a<©aoo»coooc~t''©cO'«j'ooi-H-*j<t-©e'j'«a<t-a>©e'»e*5cot- 

O©rHiHf^C'lf0'^'©t»©f<lT»(C— ©■**C-rH©a5©©e')CO©U3«-l©t—©t-C') 

iHi-H^c^c^c^cocoTfiot^ooaiiMc^^mt^o^fooo 

HHHHHHWC'I 

Diameters 

npa 

(1BJ, 

n! “5 Sn IS d< m2 "2 22 'll _W a-N -«« «>0 a « -w 

H H «n -n m* nn w ~w mo nn ®„ „„ 

puaaqx 
jo uioTjog 

rHc^oJMTfri<©©t : - o o© o ^i< : vjfCT*<©t-a5^-^©aa-<eAiaoo©oi-5.S5! 

HHHHHHHHNNMNMMMrtTjl ■*! m 

Bolt 

qom J9d 
spB9jqx 

S 2 2 ^ 2 £2 ^ 2 °* °° ** *"* ° ^*0 •'*■»* coco coco co 

soqoux 

J9X9U1BIQ 

»H ft »H CJ CO -O" 

•H’ _, »HrHCMO»COC(9^^0 




















































MACHINE DESIGN 


51 


is no friction, the equation for the equality of the work at the handle 
and at the screw is 

F2nl = Pp (13) 

Friction, however, is always present; hence the ratio of the useful 
work (Pp) to the work applied (F2nl) is not unity as above re¬ 
lations assume. From numerous experiments on the friction of screws 
and nuts, it has been found that the efficiency may be as low as 10 
per cent. Introducing the efficiency in equation 13, it may be 
written 


pp 

F2nl 


(14) 


Assuming that 50 pounds is exerted by a workman in tightening 
up the nut on a l-inch bolt, the equation above shows that P = 4,021 




Fig. 20. 


|-— P —-J 




pounds; or the initial tension is somewhat less than the tabular safe 
load shown for a 1-inch bolt, with S assumed at 10,000 pounds per 
square inch. 

For shearing stresses the bolt should be fitted so that the body 
of the bolt, not the threads, resists the force tending to shear off 
the bolt perpendicular to its axis. The internal strength of the 
bolt to resist shear is the allowable stress 5 times the area of the 

nr 

bolt in shear, or—-. If W represents the external force tending 


4 




























52 


MACHINE DESIGN 


to shear the bolt the equality of the external force to the internal 
strength is 

jy — 

4 

Reference to Table IV on Page 50 for the shearing strength of bolts, 
may be made to save the labor of calculations. 

Let Fig. 20 represent a square thread screw for the transmis¬ 
sion of motion. The surface on which the axial pressure bears, if 



n is the number of threads in the cut, is — (d 2 — d 2 )n. Suppose that 


a pressure of k pounds per square inch is allowed on the surface of 
the thread. Then the greatest permissible axial load P must not 
exceed the allowable pressure; or, equating, 

P = k -^-(d 2 - d?) n (16) 


The value of k varies with the service required. If the motion be 
slow and the lubrication very good, k may be as high as 900. For 
rapid motion and doubtful lubrication, k may not be over 200. 
Between these two extremes the designer must use his judgment, 
remembering that the higher the speed the lower is the allowable 
bearing pressure. 

Practical Modification. It will be noted in the formulas for 
bolt strengths that different values for S are assumed. This is neces¬ 
sary on account of the uncertain initial stresses which are produced 
in setting up the nuts. For cases of mere fastening, the safe tension 
is high, as just before the joint opens the tension is about equal to 
the load and yet the fastening is secure. On the other hand, bolts 
or studs fastening joints subjected to internal fluid pressure must 
be stressed initially to a greater amount than the working pressure 
which is to come on the bolt. As this initial stress is a matter of 
judgment on the part of the workman, the designer, in order to be 
on the safe side, should specify not less than f-inch or J-inch bolts 
for ordinary work, so that the bolts may not be broken off by a careless 
workman accidentally putting a greater force than necessary on the 
wrench handle. In making a steam-tight joint, the spacing of the 
bolts will generally determine their number; hence often an excess 
of bolt strength is found in joints of this character. 

Through bolts are preferred to studs, and studs to tap bolts 




MACHINE DESIGN 


53 


or cap screws. If possible, the design should be such that through 
bolts may be used. They are cheapest, are always in standard 
stock, and will resist rough usage in connecting and disconnecting. 
The threads in cast iron are weak and have a tendency to crumble; 
and if a through bolt cannot be used in such a case, a stud, which 
can be placed in position once for all, should be employed—not a 
tap bolt, which injures the thread in the casting every time it is re¬ 
moved. 

The plain portion of a stud should be screwed up tight against 
the shoulder, and the tapped hole should be deep enough to prevent 
bottoming. To avoid breaking off the stud at the shoulder, a groove 
may be made at the lower end of the thread entering the nut. 

To withstand shearing forces the bolts must be fitted so that 
no lost motion may occur, otherwise pure shearing will not be secured. 

Nuts are generally made hexagonal, but for rough work are 
often made square. The hexagonal nut allows the wrench to turn 
through a smaller angle in tightening up, and is preferred to the 
square nut. Experiments and calculations 
show that the height of the nut with stand¬ 
ard threads may be about \ the diameter of 
the bolt and still have the shearing strength 
of the thread equal to the tensile strength of 
the bolt at the root of the thread. Practi¬ 
cally, however, it is difficult to apply such a 
thin wrench as this proportion would call for 
on ordinary bolts. More commonly the height 
of the nut is made equal to the diameter of 
the bolt so that the length of thread will 
guide the nut on the bolt, give a low bearing 
pressure on the threads, and enable a suitable wrench to be easily 
applied. The standard proportions for bolts and nuts may be 
found in any handbook. Not all manufacturers conform to the 
United States standard; nor do manufacturers in all cases conform 
to one another in practice. 

If the bolt is subject to vibration, the nuts have a tendency to 
loosen. A common method of preventing this is to use double 
nuts, or loch nuts , as they are called, Fig. 21. The under nut is 
screwed tightly against the surface, and held by a wrench while 




















54 


MACHINE DESIGN 


the second nut is screwed down tightly against the first. The effect 
is to cause the threads of the upper nut to bear against the under 
sides of the threads of the bolt. The load on the bolt is sustained, 



Pig. 22. Buttress Thread 


therefore, by the upper nut, which should be the thicker of the two; 
but for convenience in applying wrenches the position of the nuts is 
often reversed. 

The form of thread adapted to transmitting power is the square 



Fig. 23. Modification of Square Thread 

thread, which, although giving less bursting pressure on the nut, is 
not as strong as the V-thread for a given length, since the total sec¬ 



tion of thread at the bottom is only \ as great. If the pressure is 
to be transmitted in but one direction, the two types may be com- 








































































MACHINE DESIGN 


55 


bined advantageously to form the buttress thread of the proportions 
shown in Fig. 22. Often, as in the carriage of a lathe, to allow the 
split nut to be opened and closed over the lead screw, the sides of the 


iI±j 



Fig. 27. Set Screws 


thread are placed at a small angle, say 15°, to each other, as illustrated 

in Fig. 23. 

The practical commercial forms of fastenings are usually in¬ 
cluded in five classes, as follows: 

1. Through bolts , Fig. 24, usually rough stock, with square 
upset heads, and square or hexagonal nuts. 

2. Tap bolts , Fig. 25, also called cap screws. These usually 
have hexagonal heads, and are found both in the rough form, and 
finished from the rolled hexagonal bar in the screw machine. 



<=>o* 


Fillister Head 




Fig. 28. Machine Screws 


3. Studs , Fig. 26, rough or finished stock threaded in the 
screw machine. 









































































56 


MACHINE DESIGN 


4. Set screws, Fig. 27, usually with square heads and case- 
hardened points. Many varieties of set screws are made, the prin¬ 
cipal distinguishing feature of each being in the shape of the point. 
Thus, in addition to the plain beveled point, there are the “cupped,” 
rounded, conical, and “teat” points. 

5. Machine screws , Fig. 28, usually fillister, round, or flat head. 
Common proportions are indicated relative to diameter of body of 
screw. 

Examples. 1 . What is the working stress on a bolt 1J inches 
in diameter if the load on the bolt is 5,500 lbs.? 

Solution . Referring to Table IV, it is found that a lj in. bolt 
has an area at the bottom of the thread of .89 sq. in. 

W 


W = Sa S = 


a 


S = 


5500 

—gg - = 6,180 lbs., approximately 


2. A wrench 15 inches long is used to screw up a nut on a 1-inch 
bolt. If the efficiency is 12 per cent, what axial load is exerted 
when a force of 25 pounds is applied at one end of the wrench handle? 
Solution. Using equation 14 

Pp 


F2rd 


= .12 


or 


.\2F2tzI 


V 

From Table IV, it is found that a 1-inch bolt has 8 threads per 
inch; then 

1 


V = 


= ~ = .125 


8 


P = .12 X 


25 X 2 X t r X 15 
.125 


= 2262.0 lbs. nearly 


PROBLEMS FOR PRACTICE 

1. Calculate the diameter of a bolt to sustain a load of 5,000 lbs. 

2. The shearing force to be resisted by each of the bolts of a 
flange coupling is 1,200 lbs. What commercial size of bolt is required? 

3. With a wrench 16 times the diameter of the bolt, and an 







MACHINE DESIGN 


57 


efficiency of 10 per cent, what axial load can a man exert on a stand¬ 
ard f-inch bolt, if he pulls 40 lbs. at the end of the wrench handle? 

4. A single, square-threaded screw of diameter 2 inches, 
lead, J inch, depth of thread J inch, length of nut 3 inches, is to be 
allowed a bearing pressure of 300 lbs. per square inch. What axial 
load can be carried? 

5. Calculate the shearing stress at the root of the thread in 
Problem 4. 


RIVETS AND RIVETED JOINTS 


NOTATION—The following notation is used throughout the chapter on Rivets 

and Riveted Joints: 


a = Net section of plate (square 
inches) 

d = Diameter of rivet (inches) 

= Number of rivets in a row 
n 2 = Total number of rivets in a joint 
n 3 = Total number of rivets in a lap 
joint and one-half the num¬ 
ber of rivets in a butt joint 
P c — Strength of joint computed 
from bearing value of plate 
(lbs.) 


P 3 = Strength of joint computed from 
shearing volume (lbs.) 

Pi — Strength of joint computed from 
tensile strength value (lbs.) 
S c — Compressive working strength 
of plate (lbs. per sq. in.) 

S B = Shearing working strength of 
the plate (lbs. per sq. in.) 

S% = Tensile working strength of the 
plate (lbs. per sq. in.) 
t = Thickness of plate (inches) 
w = Width of plate (inches) 


Analysis. A rivet is a short bar of malleable metal with a head 
at each end, thus forming a bolt having its body head and nut in one 
piece. Rivets are used to fasten together parts where the straining 
force is parallel to the surfaces, as plates, beams, and girders. From 
this it is evident that rivets are placed in shear. The parts fastened 
together may fail due to the material crushing in front of the rivet 
holes, tearing between them, the rivets shearing, or a combination 
of the above. 

When the joint is to be steam-, air-, or gas-tight, the joint must 
be tight as well as strong. This requires the rivets to be close to¬ 
gether and near the edge in order to prevent opening of the edges. 
For purely structural purposes the joint is designed for strength. 

Kinds of Joints. A lap joint is one in which the plates or bars 
joined overlap each other, Fig. 29. A butt joint is one in which the 
plates or bars that are joined butt against each other, Fig. 30. The 
thin side plates on butt joints are called cover plates; the plates or bars 


58 


MACHINE DESIGN 


that are joined are called main 'plates; and the distance between the 
centers of consecutive holes in a row of rivets is called pitch. 

Theory. When a lap joint is subjected to tension (i. e ., when P, 
Fig. 29, is a pull), and when it is subjected to compression (when P 
is a push), there is a tendency to cut or shear each rivet along the 
surface between the two plates. In butt joints with two cover plates, 
there is a tendency to cut or shear each rivet on two surfaces, Fig. 30. 
Therefore, the rivets in the lap joint are said to be in single shear; 
and those in the butt joint (two covers) are said to be in double shear. 

The shearing value of a rivet means the resistance which it can 
safely offer to forces tending to shear it on its cross-section. This 


"CT" i ? 


t 


ST 


3 —*p 


3 




* 


i i 




Fig. 29. Lap Joint 


Fig. 30. Butt Joint 


value depends on the area of the cross-section and on the working 
strength of the material. Then, since the area of the cross-section 
equals 0.7854 d 2 , the shearing strength s of one rivet is 
for single shear s = 0.7854 d 2 S a (17) 

for double shear s = 1.5708 d* S a (18) ♦ 

When a joint is subjected to tension or compression, each rivet 
presses against a part of the sides of the holes through which it 
passes. By bearing value of a plate, in this connection, is meant 
the pressure exerted by a rivet against the side of a hole in the plate, 
which the plate can safely stand. This value depends on the thick¬ 
ness of the plate, on the diameter of the rivet, and on the compressive 
working strength of the plate. Exactly how it depends on these 
three qualities is not known, but the bearing value is always com¬ 
puted from the expression tdS c . 

The holes punched or drilled in a plate or bar weaken its tensile 
strength, and to compute that strength it is necessary to allow for 
the holes. By net section, in this connection, is meant the smallest 
cross-section of the plate or bar; this is always a section along a line 
of rivet holes. Then the net section is 






















MACHINE DESIGN 


59 


a = (w — nfyt (19) 

The strength of the unriveted plate is wtS i} and the equation 
for the reduced tensile strength is 

P t = (w— n 1 d)tS i 

The compressive strength of a plate is also lessened by the 
presence of holes; but when they are again filled up, as in a joint, 
the metal is replaced, as it were, and the compressive strength of the 
plate is restored. No allowance, therefore, is made for holes in figuring 
the compressive strength of a plate. 

The strength of a joint is determined by either (1) the shearing 
value of the rivets; (2) the bearing value of the plate; or (3) the tensile 
strength of the riveted plate if the joint is in tension. Then, as before 
explained, 

P % = (w — 7i 1 d)tS i (20) 

P 8 = n 2 0.7854 (PS, (21) 

P c = n 3 tdS c (22) 

Efficiency of a Joint. The ratio of the strength of a joint to that 
of the solid plate is called the efficiency of the joint. If ultimate 
strengths are used in computing the ratio, then the efficiency is called 
ultimate efficiency; and if working strengths are used, then it is called 
working efficiency. In the following, the values refer to working 
efficiency. An efficiency is sometimes expressed as a per cent. To 

. strength of joint , AA 
express it thus, multiply the ratio, " s ~ j re ^di"of " solid " pfat e * ^ 

Practical Modification. In a butt joint the cover plates are 
made not less than one-half the thickness of the main plates. Some¬ 
times butt joints are made with only one cover plate; in such a case 
the thickness of the cover plate is never less than that of the main 
plate. 

When wide bars or plates are riveted together, the rivets are 
placed in rows, always parallel to the “seam” and sometimes also 
perpendicular to the seam; but when a row of rivets is used in this 
discussion, a row parallel to the seam is meant. A lap joint with a 
single row of rivets is said to be single-riveted; and one with two rows 
of rivets, double-riveted. A butt joint with two rows of rivets (one 
on each side of the joint) is called single-riveted, and one with four 
rows (two on each side), double-riveted. 



60 


MACHINE DESIGN 


When a joint is subjected to tension or compression, there is a 
tendency to slippage between the faces of the plates of the joint. 
This tendency is overcome wholly or in part by frictional resistance 
between the plates. The frictional resistance in a well-made joint 
may be very large, for rivets are put into a joint hot, and are headed 
or capped before being cooled. In cooling they contract, drawing 
the plates of the joint tightly against each other, and producing a 
great pressure between them, which gives the joint a correspondingly 
large frictional strength. It is the opinion of some that all well- 
made joints perform their service by means of their frictional strength; 
that is to say, the rivets act only by pressing the plates together and 
are not under shearing stress, nor are the plates under compression 
at the sides of their holes. The frictional strength of a joint, how¬ 



ever, is usually regarded as uncertain, and generally no allowance 
is made for friction in computations on the strength of riveted joints. 

To make sure of a good tight joint, it is closed up by burring 
down the edges of the plate. This is called calkmg. The calking 
tool, Fig. 31, resembles a chisel, except that the point is flat. The 
tool is forced into the plate by hand hammering and must be skill¬ 
fully done in order to prevent injury to the plates. Sometimes a 
calking tool has a thickness equal to that of the plate, Fig. 32. 

Examples. 1. Two half-inch plates, 7J inches wide, are con¬ 
nected by a single lap joint double-riveted, six rivets in two rows. 
If the diameter of the rivets is | inch, and the working strengths are 
S t = 12,000, S 8 = 7,500, and S 0 = 15,000 pounds per square inch, 
what is the safe tension which the joint can transmit? 

Solution. 

n t = 3, n 2 = 6, and n 9 — 6 














MACHINE DESIGN 


61 


Pt = P 2 — (3 X f)] X I X 12,000 = 31,500 pounds 
P e = 6 X 0.7854 X (|) 2 X 7,500 = 19,880 pounds 
P 0 = 6 X f X f X 15,000 = 33,750 pounds 

Since P s is the least of these three values, it determines the 
strength of the joint, viz, 19,880 pounds. 

2. Suppose that the plates described in the preceding example 
are joined by means of a butt joint (two cover plates), and 12 
rivets are used, being spaced as before. What is the safe tension 
which the joint can bear? 

Solution. Here n t = 3, n 2 =12, and n 3 = 6; hence, as in the 
preceding example, 

P t = 31,500; andP c = 33,750 pounds; but 

P s = 12 X 0.7854 X (f) 2 X 7,500 = 39,760 pounds 

The strength equals 31,500 pounds, and the joint is stronger than 
the first. 

3. Suppose that in the preceding example the rivets are ar¬ 
ranged in rows of two. What is the tensile strength of the joint? 

Solution. Here n x = 2, n 2 = 12, and n 3 = 6; hence, as in the 
preceding example, 

P 8 = 39,760; and P c = 33,750 pounds; but 

P t = [7b - (2 X })] X i X 12,000 = 36,000 pounds 

The strength equals 33,750 pounds, and this joint is stronger than 
either of the first two. 

PROBLEMS FOR PRACTICE 

St = 12,000, S s = 7,500, and S c = 15,000 pounds per square inch. 

1. Two half-inch plates, 5 inches wide, are connected by a lap 
joint, with two f-inch rivets in a row. What is the safe strength of 
the joint? 

2. Solve the preceding example supposing that four f-inch 
rivets are used, in two rows. 

3. Solve Example 1 assuming three 1-inch rivets placed in a 
row lengthwise of the joint. 

4. Two half-inch plates, 5 inches wide, are connected by a butt 
joint (two cover plates), and four f-inch rivets are used, in two rows. 
What is the strength of the joint? 


62 


MACHINE DESIGN 


5. It is required to compute the efficiencies of the joints de¬ 
scribed in the illustrative examples. 

In each case the plate is \ inch thick and 7| inches wide; hence 
the tensile working strength of the solid plate is 

1\ X \ X 12,000 = 45,000 pounds 


KEYS, PINS, AND COTTERS 


NOTATION—The following notation 

Keys, Pins, 

D = Average diameter of rod (inches) 
Z>!= Outside diameter of socket (in¬ 
ches) 

d — Diameter of shaft (inches) 

L — Length of key (inches) 

P = Driving force (lbs.) 

Pj= Axial load on rod (lbs.) 

R = Radius at which P acts (inches) 

S = Safe crushing fiber stress (lbs. 
per sq. in.) 


is used throughout the chapter on 
and Cotters: 


S, 


St = 

T = 
W = 
w = 

Wy = 

Wo = 


= Safe shearing fiber stress (lbs. 
per sq. in.) 

Safe tensile fiber stress (lbs. per 
sq. in.) 

Thickness of key or cotter (inches) 
Width of key (inches) 

Average width of cotter (inches) 
End of slot to end of rod (inches) 
End of slot to end of socket (in¬ 
ches) 


KEYS AND PINS 


Analysis. Keys and pins are used to prevent relative rotary 
motion between machine parts intended to act together as one piece. 
If a hole is drilled completely through a hub and across the shaft, 
and a tightly fitted pin is inserted, any rotary motion of the one will 
be transmitted to the other, provided the pin does not fail by shear¬ 
ing off at the joint between the shaft and the hub. The shearing 
area is the sum of the cross-sections of the pin at the joint. 

A hole may be drilled in the joint, the axis of the hole being 
parallel to the axis of the shaft, and a pin may be driven in introducing 
a shearing area as before, but the area is now equal to the diameter 
of the pin multiplied by its length, and the pin is stressed sidewise, 
instead of across. It is evident in the sidewise case that the shearing 
area may be increased to anything desired without changing the 
diameter of the pin, merely by increasing the length of the pin. 

As there are some manufacturing reasons why a round pin 
placed lengthwise in the joint is not always applicable, a rectangular 
pin may be used, in which case it is called a key. 

When pins are driven across the shaft, as in the first instance, 
they are usually made taper. This is because it is easier to ream 
a taper hole to size than a straight hole, and a taper pin will drive 


MACHINE DESIGN 


63 


more easily than a straight pin, it not being necessary to match the 
hole in hub and shaft so exactly in order that the pin may enter. 
The taper pin will draw the holes into line as it is driven, and can 
be backed out readily in removal. 

Keys of the rectangular form are either straight or tapered, 
but for different reasons from those just stated for pins. Straight 
keys have working bearing only at the sides, driving purely by shear, 
crushing being exerted by the side of the key in both shaft and hub, 
over the area against the key. The key itself does not prevent end 
motion along the shaft; and if end motion is not desired, auxiliary 
means of some sort must be resorted to, as, for example, set screws 
through the hub jamming hard against the top of the key. 

If end motion along the shaft is desired, the key is called a spline , 
and, while not jammed against the shaft, is yet prevented from 



changing its relation to the hub by some means such as illustrated 
in Fig. 33. 

Taper keys not only drive through sidewise shearing strength, 
but prevent endwise motion by the wedging action exerted between 
the shaft and hub. These keys drive more like a strut from corner 
to corner; but this action is incidental rather than intentional, and the 
proportions of a taper key should be such that it will give its full 
resisting area in shearing and crushing, the same as a straight key. 

Theory. Suppose that the pin illustrated in Fig. 34 passes 
through hub and shaft, and the driving force P acts at the radius 











































64 


MACHINE DESIGN 


R; then the force which is exerted at the surface of the shaft to 

9 pp 

shear off the pin at the points A and B is — -j— . If D t is the 

average diameter of the pin, its shearing strength is w n ^ 

Equating the external force to the internal strength 

2 PR 2 7i D t 2 S a 
d 4 


or 



In Fig. 35 a rectangular key is sunk half way in hub and shaft 
according to usual practice. Here the force at the surface of the 



shaft, calculated the same as before, not only tends to shear off the 
key along the line AB y but tends to crush both the portion in the 
shaft and in the hub. The shearing strength along the line AB 
is LWS V Equating external force to internal strength 


2 PR 
d 


= LWS, 


* 


or 


W = 


2 PR 
dLS, 


( 24 ) 


The crushing strength is, of course, that due to the weaker 
metal, whether in shaft or hub. Let S c be this least safe crushing 

LT 

fiber stress. The crushing strength then is —L- S e , and, equating 
external force to internal strength, 

































MACHINE DESIGN 


t'O 


or 


2 PR LT c 

d ~ 2 




The proportions of the key must be such that the equations as above, 
both for shearing and for crushing, shall be satisfied. 

Practical Modification. Pins across the shaft can be used to 
drive light work only, for the shearing area cannot be very large. 
A large pin cuts away too much 
area of the shaft, decreasing the 
latter’s strength. Pins are useful in 
preventing end motion, but in this 
case are expected to take no shear, 
and may be of small diameter. 

The common split pin is especially 
adapted to this service, and is a 
standard commercial article. 

Taper pins are usually listed 
according to the Morse standard 
taper, proportions of which may 
be found in any handbook. It 
is desirable to use standard taper pins in machine construction, as 
the reamers are a commercial article of accepted value, and readily 
obtainable in the machine-tool market. 

With properly fitted keys, the shearing strength is usually the 
controlling element. For shafts of ordinary size, the standard 
proportions as given in Tables V and VI are safe enough without 
calculation, up to the limit of torsional strength of the shaft. For 
special cases of short hubs or heavy loads, a calculation is needed 
to check the size, and perhaps modify it. 

Splines, also known as feather keys , require thickness greater 
than regular keys, on account of the sliding at the sides. Suggested 
proportions for splines are given in Table VI. 

Though the spline may be either in the shaft or hub, it is generally 
dovetailed , Fig. 36, gibbed, or otherwise fastened in the hub, and a 
long spline way made in the shaft, in which it slides. 

The straight key, accurately fitted, is the most desirable fasten- 



Fig. 36. Gibbed Key 












66 


MACHINE DESIGN 


ing device for accurate machines, such as machine tools, on account 
of the fact that there is absolutely no radial force exerted to throw 
the parts out of true. It, however, requires a tight fit of hub to shaft, 
as the key cannot be relied upon to take up any looseness. 

The taper key, Fig. 37, by its wedging action, will take up 
some looseness, but in so doing throws the parts out slightly. Or, 


k 

St¬ 



1 


ftfii_ 

J" 

s Taper /s Per Ft. 


A 




Fig. 37. Taper Key 



even if the bored fit is good, if the taper key is not driven home 
with care, it will spring the hub, and make the parts run untrue. 
The great advantage, however, that the taper key has of holding 
the hub from endwise motion, renders it a very useful and practical 
article. It is usually provided with a head, or gib, wdiich permits 
a draw hook to be used to wedge between the face of the hub and the 
key to facilitate starting the key from its seat. 

Two keys at 90° from each other may be used in cases where 
one key will not suffice. The fine workmanship involved in spacing 
these keys so that they will drive equally makes this plan inadvisable 
except in case of positive and unavoidable necessity. 

The “Woodruff” key, Fig. 38, is a useful patented article for 



certain locations. This key is a half-disk sunk in the shaft and the hub 

is slipped over it. A simple rotary cutter is dropped into the shaft 
to produce the key seat; because of the depth in the shaft, the 
tendency to rock sidewise is eliminated, and the drive is by shear. 















































MACHINE DESIGN 


67 


TABLE V 

Proportions for Gib Keys 


Diameter of shaft ( d ) inches 

a 

4 

1 


1 5 

2 

2} 

31 

4 

5 

6* 

Width 

(TV) inches 

X 

1 

X 

i 

% 

% 

1 

IX 

ix 

If 

Thickness 

( T ) inches 

i 

4 


X 

ft 

X 

1 7 

¥2 

2 1 

3 2 

X 

1 

n 


Keys may be milled out of solid stock, or drop-forged to within 
a small fraction of finished size. The drop-forged key is an excellent 
modern production and requires but a minimum amount of fitting. 
Any key, no matter how produced, requires some hand fitting and 
draw filing to bring it properly to its seat and give it full bearing. 

It is good mechanical policy to avoid keyed fastenings when¬ 
ever possible. This does not mean that keys may never be used, 
but that a key is not an ideal way to produce an absolutely positive 
drive, partly because it is an expensive device, and partly because 
the tendency of any key is to work itself loose, even if carefully fitted. 

Tables V and VI are suggested as a guide to proportions of 
gib keys and feather keys, and will be found useful in the absence 
of any manufacturer’s standard list. 


TABLB VI 

Proportions for Feather Keys 


Diameter of shaft ( d ) inches 

f 

1 


H 

2 

2i 

2i 

3 

3i 

4 

H 

Width 

( W ) inches 

X 

S'2 

i 

4 

\ 

X 

a 

8 

1 

1 

2 

X 

X 

A 

8 

Thickness 

( T) inches 

1 

X 

f 

t 

X 

1 

2 


f 

I 

I 

1 

S 


COTTERS 

Analysis. Cotters are used to fasten hubs to rods rather than 
to shafts, the distinction between a rod and a shaft being that a rod 
takes its load in the direction of its length and does not dri\e b\ 
rotation. A cotter, therefore, is nothing but a cross-pin of modified 
form, to take shearing and crushing stress in the direction of the axis 

of the rod, instead of perpendicular to it. 

Referring to Fig. 29, one will see that the cottei is made long 
and thin—long, in order to get sufficient shearing area to resist 
shearing along lines A and B; thin, in order to cut as little cross- 
sectional area out of the body of the shalt as possible. The cottei 
































68 


MACHINE DESIGN 


itself tends to shear along the lines A and B , and crush along the 
surfaces K , G, and J. The socket tends to crush along the surfaces 
K and G. The rod end D tends to be sheared out along the lines 
C II and Q E, and also to be crushed along the surface J. The 
socket tends to be sheared along the lines V U and X Y. 

The cotter is made taper on one side, thus enabling it to draw 
up the flange of the rod tightly against the head of the socket. 



This taper must not be great enough to permit easy backing out 
and loosening of the cotter under load or vibration in the rod. In 
responsible situations this loosening cannot be safely guarded against 
except through some auxiliary locking device, such as lock nuts on 
the end of the cotter, Fig. 40. 

Theory. Referring to Fig. 39, assume an axial load of P 
as shown. The successive equations of external force to internal 
strength are enumerated below, for the different actions that take 
place. 







































































MACHINE DESIGN 


69 


For shearing along lines A and B 

P x = 2 TwS b (26) 

in which w is the average width of cotter, and S 9 its safe shearing 
stress. 

For crushing along surfaces K and G 

= T (D 1 - D) S c (27) 

in which S c is least safe crushing stress, whether of cotter or socket. 
For crushing along surface J 

P t = DTS C (28) 

in which S c is least safe crushing stress, whether of cotter or socket. 
For shearing along surfaces CH and QE 

P j = 2 w 1 DS s (29) 

in which S t is the safe shearing stress of rod end, and w t end of slot 
to end of rod. 

For tension in rod end at section across slot 

P t Td)s, (30) 

in which is safe tensile stress in rod end. 

For tension in socket at section across slot 

[ ; zD 2 7 cD 2 1 

- T (D t - D)\s t (31) 

in which S t is safe tensile stress in socket. 

For shearing in socket along the lines VU and XY 

P, = 2 w 2 ( D , - D) S, (32) 

in which S B is safe shearing stress in the socket, and w 2 end of slot 
to end of socket. 

The proportions of cotter and socket may be fixed to some 
extent by practical or assumed conditions. The dimensions may 
then be tested by the above equations, that the safe working stresses 
may not be exceeded, the dimensions being then modified accordingly. 

The steel of which both cotter and rod would ordinarily be made 
has range of working fiber stress as follows: 

Tension, 8,000 to 12,000 (lbs. per sq. in.). 

Compression, 10,000 to 16,000 (lbs. per sq. in.). 

Shear, 6,000 to 10,000 (lbs. per sq. in.). 

The socket, if made of cast iron, will be weak as regards tension, 





70 


MACHINE DESIGN 



Fig. 40. Cotter with Locking Device 


tendency to shear out at the end, and tendency to split. The uncer¬ 
tainty of cast iron to resist these is so great that the hub or socket 

must be very clumsy in order to 
have enough surplus strength. This 
is always a noticeable feature of 
the cotter type of fastening, and 
cannot well be avoided. 

Practical Modification. The 
driving faces of the cotter are 
often made semicircular. This 
not only gives more shearing area 
at the sides of the slots, but makes 
the production of the slots easier 
in the shop. It also avoids the 
general objection to sharp corners 
—namely, a tendency to start 
cracks. 

A practicable taper for cotters 
is \ inch per foot. This will, 
under ordinary circumstances, prevent the cotter from backing out 
under the action of the load. 

When set screws or lock nuts 
are used, as in Fig. 40, the 
taper may be greater than 
this, perhaps as much as 1J 
inches per foot. 

In the common use of 
the cotter for holding the 
strap at the ends of connect¬ 
ing rods, the strap acts like a 
modified form of socket. 

This is shown in Figs. 40 
and 41. Here, in addition 
to holding the strap and rod 
together lengthwise, it may be necessary to prevent their spreading, 
and for this purpose an auxiliary piece G with gib ends is used. 
The tendency without this extra piece is shown by the dotted 
lines in Fig. 41. 



Fig. 41. Cotter at End of Connecting Hod 



















































MACHINE DESIGN 


71 


The general mechanical fault with cottered joints is that the 
action of the load, especially when it constantly reverses, as in pump 
piston rods, always tends to work the cotter loose. Vibration also 
tends to produce the same effect. Once this looseness is started in 
the joint, the cotter loses its pure crushing and shearing action, and 
begins to partake of the nature of a hammer, and pounds itself and 
its bearing surfaces out of their true shape. Instead of a collar on 
the rod, a taper fit of the rod in the socket is often found; and any 
looseness in this case is still worse, for the rod then has end play in 
the socket, and by its shucking back and forth tends to split open the 
socket. 

The only answer to these objections is to provide a positive 
locking device, and take up any looseness the instant it appears. 

Examples. 1 . A gear 60 inches in diameter has a load of 
3,000 pounds at the pitch line. The shaft is 4 inches in diameter 
in a hub 5 inches long; and the key is a standard gib key as given 
in the table. Test its proportions for shearing. 


Solution. 


W = 


2 PR 


dLS Q 


W = 1 


16 


T = 


13 

16 


1 _ 2 X 3000 X 30 
] l6 “ 4 X 5 X S a 


2 X 3000 X 30 
” 4 X 5 X 1.0625 

= 8450 lbs. per sq. in. 


9,000 is about the limit for safety. The key is, therefore, safe for 
shearing stresses. 

2. A piston rod 2 inches in diameter carries a cotter f inch 
thick, and has an axial load of 20,000 pounds. Calculate the 
average width of the cotter. S s = 9,000. 

Solution. P= 2 TwS, 20000 = 2 X f w X 9000 


w = 


2 X 


20000 

.375 X 9000 


= 2.97, approximately 3 inches 









72 


MACHINE DESIGN 


PROBLEMS FOR PRACTICE 

1. Calculate the safe load in shear which can be carried on a 
key \ inch wide, f inch thick, and 5 inches long. Assume S a = 6,000. 

2. Assuming the above key to be y 8 ^ inch in hub and r 8 y inch 
in shaft, test its proportions for crushing at S c = 16,000. 

3. A piston rod 2 inches in diameter carries a cotter f inch, 
and has an axial load of 20,000 pounds. How far from the end 
of the rod must the end of the slot be? 

4. Calculate the fiber stress in the rod in the preceding prob¬ 
lem at a section through the slot. 

COUPLINGS 

NOTATION—The following notation is used throughout the chapter on 

Couplings: 

D = Diameter of shaft (inches) S c = Safe crushing fiber-stress (lbs. 

d = Diameter of bolt body (inches) per sq. in.) 

n — Number of bolts T = Twisting moment (inch-lbs.) 

R = Radius of bolt circle (inches) t = Thickness of flange (inches) 

S = Safe shearing fiber stress (lbs. W = Load on bolts (lbs.) 

per sq. in.) 

Analysis. Rigid couplings are intended to make the shafts which 
they connect act as a solid, continuous shaft. In order that the 
shaft may be worked up to its full strength capacity, the coupling 
must be as strong in all respects as the shaft, or, in other words, it 
must transmit the same torsional moment. In the analysis of the 
forces which come upon these couplings, it is not considered that 
they are to take any side load, but that they are to act purely as tor¬ 
sional elements. It is doubtless true that in many cases they do 
have to provide some side strength and stiffness, but this is not their 
natural function, nor the one upon which their design is based. 

Referring to Fig. 42, which is the type most convenient for 
analysis, the simplest form of flange coupling is shown. It consists 
merely of hubs keyed to the two portions, with flanges driving through 
shear on a series of bolts arranged concentrically about the shaft. 
The hubs, keys, and flanges are subject to the same conditions of 
design as the hubs, keys, and web of a gear or pulley, the key tend¬ 
ing to shear and be crushed in the hub and shaft, and the hub tend¬ 
ing to be torn or sheared from the flange. The driving bolts, which 
must be carefully fitted in reamed holes, are subject to a purely 


MACHINE DESIGN 


73 


shearing stress over their full area at the joint, and at the same time 
tend to crush the metal in the flange, against which they bear, over 
their projected area. This latter stress is seldom of importance, the 
thickness of the flange, for practical reasons, being sufficient to make 
the crushing stress very low. 

Theory. The theory of couplings, being the same as for keys, 
need not be repeated. The shearing stress on the bolts is the only 
new point to be studied. 

In Fig. 42, for a twisting moment on the shaft of T , the 



Fig. 42. Simple Flange Coupling 


T 

load at the bolt circle is W = —. If the number of bolts be ?z, 

R 

equating the external force to the internal strength 

U7 T Snd 2 ,oq\ 

W = ™ (33) 

R 4 

Although the crushing stress will seldom be used, yet for the 
sake of completeness its equation is given. 

W = -- = S.dtn (34) 

R 

SD 3 

The internal moment of resistance of the shaft is —hence 

5.1 

the equation representing full equality of strength between the shaft 
and the coupling, depending upon the shearing strength of the bolts, is 

SD 3 Snd 2 

n 


5.1 R 


4 


(35) 


























































74 


MACHINE DESIGN 


The theory of the other types of couplings is obscure, except 
as regards the proportions of the key, which are the same in all 
cases. The shell of the clamp coupling, Fig. 43, should be thick 
enough to give equal torsional strength with the shaft; but the exact 
function which the bolts perform is difficult to determine. In gen¬ 
eral the bolts clamp the coupling tightly on the shaft and provide 
rigidity, but the key does the principal amount of the driving. The 
bolt sizes, in these couplings, are based on judgment and relation to 
surrounding parts, rather than on theory. 

Practical Modification. All couplings must be made with care 
and nicely fitted, for their tendency, otherwise, is to spring the shafts 
out of line. In the case of the flange coupling, the two halves may 
be keyed in place on the shafts, the latter then swung on centers in 



Fig. 43. Clamp Coupling 



the lathe, and the joint faced off. Thus the joint will be true to the 
axis of the shaft; and, when it is clamped in position by the bolts, 
no springing out of line can take place. 

A flange F, Fig. 42, is sometimes made on this form of coupling, 
in order to guard the bolts. It may be used, also, to take a light 
belt for driving machinery; but a side load is thereby thrown on the 
shaft at the joint, which is at the very point where it is desirable to 
avoid it. 

The simplest form of rigid coupling, known as the muff coupling, 
consists of a plain sleeve slipped over from one shaft to the other, 
when the second is butted up against the first. This is a very satis¬ 
factory form of coupling, as it is perfectly smooth on the outside, and 
consists of only a sleeve and a key. It is, however, expensive to fit, 
difficult to remove, and requires an extra space of half its length on 
the shaft over which to be slipped back. 






























































MACHINE DESIGN 


75 


Where the flange coupling, Fig. 42, would be unnecessarily 
expensive, the clamp coupling , Fig. 43, which is simply a muff coup¬ 
ling split in halves and recessed for bolts, is a good coupling for 
moderate-sized shafts. It is cheap and it is easily applied and 
removed, even with a crowded shaft. If bored with a piece of paper 
in the joint, when it is clamped in position it will pinch the shaft 



tightly and make a rigid connection. It is desirable to have the bolt- 
heads protected as much as possible, and this may be accomplished 
by making the outside diameter large enough so that the bolts will 
not project. Often an additional shell is provided to encase the 
coupling completely after it is located. 

There are many other special forms of couplings, some of them 



Fig. 45. Oldham Coupling 


adjustable; most of them depending upon a wedging action exerted 
by taper cones, screws, or keys. Trade catalogues are to be sought 
for their description. 

The claw coupling, Fig. 44, is nothing but a heavy flange coup¬ 
ling with interlocking claws or jaws on the faces of the flanges to 





















































































76 


MACHINE DESIGN 


take the place of the driving bolts. This coupling can be thrown in 
or out as desired, although it usually performs the service of a rigid 
coupling, as it is not suited to clutching-in during rapid motion, 
like a friction clutch. 

Flexible couplings, which allow’ slight lack of alignment, are 
made by introducing between the flanges of a coupling a flexible 
disk, the one flange being fastened to the inner circle of the disk, 
the other to the outer circle. This is also accomplished by pro¬ 
viding the faces of the flange coupling with pins that drive by pres¬ 
sure together or through leather straps wrapped round the pins. 
These devices are mostly of a special and often uncertain nature, 
lacking the positiveness which is one essential feature of a good 
coupling. 

Oldham Coupling. Fig. 45 shows a form of coupling used when 
two shafts are parallel but not in line. A disk is keyed on the end 

of each shaft. Between these 
disks lies a third which has a 
feather on each side and at right 
angles to each other and fitting 
in a slot in the corresponding 
disk. The middle disk revolves 
around an axis parallel to the 
shafts and midway between them. 
The sliding of the feathers in the 
slots allows for the lack of align¬ 
ment of the shafts. Since the 
feathers are at right angles to 
each other, the slots are held at 
right angles by the disk and both 
shafts must turn with the same angular velocity. 

Universal Couplings. In case tw’o shafts are not in line and 
the angle between them is less than 45° they may be connected by a 
universal coupling, as shown in Fig. 46. There is an objection to this 
coupling, and that is that the velocity ratio of the two shafts is a con¬ 
stantly varying one, although they make quarter revolutions in equal 
times. The change in velocity ratio varies with the angle between 
the shafts, increasing as it increases. 

Example. A flange coupling of the type shown in Fig. 42 is 



mr 

Fig. 46. Universal Coupling 




















































MACHINE DESIGN 


77 


used on a shaft 2 inches in diameter. The hub is 3 inches long and 
carries a standard key. The bolt circle is 7 inches in diameter, and 
it is desired to use f-inch bolts. How many bolts are needed to trans¬ 
mit 60,000 inch-pounds for a fiber stress in the bolt of 6,000? 

Solution. 

R = = 3.5 in. T = 60000 

A 

T _ 60000 
R~ 3.5 


Shearing area of a f-in. bolt 


3.1416 


X(f) j 


Total shearing area of n f-in. bolts = 


3.1416 X 25 X 
4 X 8 2 


W = 


3.1416 X 25 X n X 6000 
4 X 64 


60000 3.1416 X 25 X 6000 X n 

3.5 4 X 64 

_ 60000 X 4 X 64 

3.5 X 3.1416 X 25 X 6000 ' 

Ans. 10 bolts. 


PROBLBMS FOR PRACTICE 

1. If 6 bolts were used in the above example, what diameter of 
bolt would be required? 

2. If four J-inch bolts were used on a circle of 8 inches diameter, 
what diameter of shaft would be used in the coupling to give equal 
strength with the bolts? 


FRICTION CLUTCHES 

NOTATION—The following notation is used throughout the chapter on 

Friction Clutches: 

a = Angle between clutch face and R = Mean radius of friction surface 
axis of shaft (degrees) (inches) 

H = Horse-power (33,000 ft.-lbs. per T = Twisting moment about shaft 
minute) axis (inch-lbs.) 

/i = Coefficient of friction (per cent) V = Force normal to clutch face (lbs.) 

N = Number of revolutions per min- W = Load at mean radius of friction 
ute surface (lbs.) 

P = Force to hold clutch in gear to 
produce W (lbs.) 












78 


MACHINE DESIGN 


Analysis. The friction clutch is a device for connecting at will 
two separate pieces of shaft, thereby transmitting power from the 
driving shaft to the auxiliary shaft up to the full capacity of the clutch. 



The connection is usually accomplished while the driving shaft is 
under full speed, the slipping between the surfaces which occurs 
during the throwing-in of the clutch, permitting the driven shaft 
to pick up gradually the speed of the other. The disconnection is 
made in the same manner, the amount of slipping which occurs de¬ 
pending on the suddenness with which the clutch is thrown out. 

The force of friction is the sole driving element, hence the prob¬ 
lem is to secure as large a force of friction as possible by producing 



Fig. 48. Cone Friction Clutch 


a heavy normal pressure between surfaces having a high coefficient 
of friction between them. The many varieties of friction clutches 
which are on the market or designed for some special purpose, are 



























































































MACHINE DESIGN 


79 


all devices for accomplishing one and the same effect, viz, the pro¬ 
duction of a heavy normal force or pressure between surfaces at such 
a radius from the driven axis, that the product of the force of friction 
thereby created and the radius shall equal the desired twisting moment 
about that axis. Three typical methods are shown in Figs. 47, 48, 
and 49. These drawings are not worked out in operative detail, 
but merely illustrate the principle, and are drawn in their simplest 
form. 

In Fig. 47 the normal pressure is created in the simplest pos¬ 
sible way, an absolutely direct push being exerted between the disks, 
due to the thrust P of the clutch fork. 

By taking advantage of the wedge action of the inclined faces, 
Fig. 48, a less thrust P will produce the required normal pressure at 
the radius R. 

In Fig. 49 the inclination of the faces is carried so far that the 



angle a of Fig. 48 has become zero; and by the toggle-joint action of 
the link pivoted to the clutch collar, the normal force produced may 
be very great for a slight thrust P. By careful adjustment of the 
length of the link so that the jaw takes hold of the clutch surface, 
when the link stands nearly vertical, a very easy operating device 
is secured, and the thrust P is made a minimum. 

Theory. In order to calculate the twisting moment, it must be 
remembered that the force of friction between two surfaces, Fig. 
47, is equal to the normal pressure times the coefficient of friction. 
This, in the form of an equation, using the symbols of the figure, is 

W = jxP (36) 


















































80 


MACHINE DESIGN 


Hence a force of magnitude pP may be considered as acting at the 
mean radius R of the clutch surface. The twisting moment will then be 

T = WR = fiPR (37) 


Referring to the equation which gives twisting moment in terms of 
horse-power, and putting the two expressions equal to each other 


or 


T = 630 ^ 5H = /tPR 


„ [iNPR 
63025 



This expression gives at once the horse-power that the clutch will 
transmit with a given end thrust P. 

In Fig. 48 the equilibrium of the forces is shown in the little 
sketch at the left of the figure. The clutch faces are supposed to 
be in gear, and the extra force necessary to slide the two together is 
not considered, as it is of small importance. The static equations 
then are 

W = pV 
T = WR = pVR 


_ 6302577 
N 

H _ fiVRN 
63025 


pVR 


V may be determined by using a parallelogram of forces as 
shown in the upper left-hand corner of Fig. 48. 

In Fig. 49, P would of course be variable, depending on the 
inclination of the little link. The amount of horse-power which 
this clutch would transmit would be the same as in the case of the 
device illustrated in Fig. 48, for an equal normal force V produced. 

The further theoretical design of such clutches should be in 
accordance with the same principles as for arms and webs of pulleys, 
gears, etc. The length of the hubs must be liberal in order to pre¬ 
vent tipping on the shaft as a result of uneven wear. The end 
thrust is apt to be considerable; and extra side stiffness must be pro- 






MACHINE DESIGN 


81 


vided, as well as a rim that will not spring under the radial 
pressure. 

Practical Modification. It is desirable to make the most com¬ 
plicated part of a friction clutch the driven part, for then the mech¬ 
anism requiring the closest attention and adjustment may be brought 
to rest and kept in this condition when no transmission of power 
is desired. 

Simplicity is an important practical requirement in clutches. 
The wearing surfaces are subjected to severe usage; and it is essential 
that they be made not only strong in the first place, but also capable 
of being readily replaced when worn out, as they are sure to be after 
some service. 

Of the three forms of clutch shown, the one in Fig. 49 is the 
most efficient, although its commercial design is considerably dif¬ 
ferent from that indicated. Usually the jaws grip both sides of 
the rim, pinching it between them. This relieves the clutch rim of 
the radial unbalanced thrust. Adjusting screws for taking up the 
wear in the jaws, and lock nuts for maintaining their position must 
be provided. 

Theoretically, the rubbing surfaces should be of those materials 
whose coefficient of friction is the highest; but the practical ques¬ 
tion of wear comes in, and hence usually both surfaces are made of 
metal, cast iron being most common. For metal on metal the 
coefficient of friction jj. cannot be safely assumed at more than 15 
per cent, because the surfaces are sure to get oily. 

A leather facing on one of the surfaces gives good results as 
to coefficient of friction, \i having a value, even for oily leather, of 
20 per cent. Much slipping, however, is apt to burn the leather; 
and this is most likely to occur at high speeds. 

Wood on cast iron gives a little higher coefficient of friction 
for an oily surface than metal on metal. Wood blocks can be so 
set in the face of the jaws as to be readily replaced when worn, and 
in such case make an excellent facing. 

The angle a of a cone friction clutch, Fig. 48, may evidently 
be made so small that the two parts will wedge together tightly with 
a very slight pressure P; or it may be so large as to have little wedging 
action, and approach the condition illustrated in Fig. 47. Between 
these limits there is a practical value which neither gives a wedging 


82 


MACHINE DESIGN 


action so great as to make the surfaces difficult to pull apart, nor, 
on the other hand, requires an objectionable end thrust along the 
shaft in order to make the clutch drive properly. 

For a = about 15 °, the surfaces will free themselves when P is relieved. 
“ a = “ 12°, “ “ require slight pull to be freed. 

“ a — “ 10°, “ “ cannot be freed by direct pull of the 

hand, but require some leverage to produce the necessary force P. 

Example. What force must be exerted to hold in a friction 
clutch for transmitting 30 horse-power at 200 revolutions per minute, 
assuming working radius of clutch to be 12 inches; coefficient of 
friction 15 per cent; angle a = 15°? 

Solution. 

y _ 6302 oH 
[lRN 

v _ 63025 X 30 

” .15 X 12 X 200 

I _ 5^2 _ 2020 p 0un( j s 

— £ 

Using parallelogram of forces P = 1313 pounds 

PROBLEMS FOR PRACTICB 

1. With what force must a friction clutch of the form shown 
in Fig. 47 be held in, in order to transmit 20 horse-power? The 
speed to be 150 revolutions per minute, assuming the radius of the 
clutch to be 10 inches and coefficient of friction 15 per cent. 

2. What horse-power could be transmitted if the working 
radius were decreased to 8 inches? 



LEATHER BELTS 

NOTATION—The following notation is used throughout the chapter on 

Belts: 


A = Sectional area of belt (square 
inches) = bh 

b = Width of belt (inches) 

F — Force of friction at pulley rim 
(lbs.) 

h = Thickness of belt (inches) 

N = Number of revolutions of pulley 
per minute 

P = Driving force at pulley rim (lbs.) 
= F 


R = Radius of pulley (feet) 
r = Radius of pulley (inches) 

T = Initial tension (lbs.) 

T n = Total tension on tight side (lbs.) 
T 0 = Total tension on slack side (lbs.) 
V = Velocity of belt (feet per min¬ 
ute) 

w = Weight of belt per cubic inch 
(lbs.) 






MACHINE DESIGN 


83 


Analysis. When a belt stretched over a pair of pulleys is cut 
off at the proper length, and is laced together into an endless band, 
it is evident that as long as the belt is at rest there is a nearly uniform 
tension in it throughout its length, due to the tightness with which 
the lacing is drawn up. If the distance between the pulleys is con¬ 
siderable, the weight of the belt itself as it hangs between the pulleys 
will produce a slightly greater tension next to the pulleys than exists 
in the middle of the span. This increase of tension due to the weight 
of the belt would make but little difference in the unit-stress in the 
material of which the belt is made; hence it may safely be assumed 
that the tension in the belt when at rest is uniform throughout its 
entire length. 

When by turning one of the pulleys, power is transmitted through 
the belt to the other pulley, the condition of stress in the belt is at 
once materially changed. The driving pulley can exert only a pull 
on the other pulley as the belt is a flexible member; the push, which 
is at the same time given to the other side of the belt, tends to make 
it sag or become slack. Hence, the immediate effect of the starting 
motion in a belt is to change the condition of equal tension to that of 
unequal tension in the two sides, the driving side being tight, and 
the other loose. The former has gained as much tension as the latter 
has lost, and the sum of the two is practically equal to the sum of 
the tensions in the two sides of the belt when at rest. This is not 
strictly true, as will be shown later, but is sufficiently accurate to 
form a good basis for the practical design, at least, of slow-speed belts. 

The possibility of transmission of power is due, of course, to the 
friction existing between the belt and the pulleys, and the amount of 
pull that may be applied to the belt is, therefore, limited by the ten¬ 
sion at which the belt slips around the pulley. Moreover, since the 
force of friction between the belt and the pulley is dependent upon 
the normal force with which the belt is pressed against the pulley, 
and the coefficient of friction between the two, it is evident that the 
tighter the belt is laced up, and the rougher the surfaces of the pulley 
and belt, the greater is the force that can be transmitted through 
the belt. This leads to the conclusion that it would be possible to 
transmit any amount of power through any belt, however small, if the 
belt were only laced up tight enough. 

The above conclusion is literally true; but the important fact 


84 


MACHINE DESIGN 


now comes in, that the strength of the material of which the belt is 
made is limited, and while theoretically it might be possible to accom¬ 
plish this, practically it would be impossible, for at a certain point 
the belt would break under the strain. Other practical considera¬ 
tions also come in, which fix this limit of power transmission at a 
point far below the breaking strength of the material. 

T^e complete analysis of the tension problem in belts is not 
quite as simple as the above, especially for high-speed belts. When 
the driving side of the belt becomes tight, it stretches and grows 
longer; and at the same time the other side of the belt becomes slack 
and grows shorter. It is not true, however, that the increase in the 

one side is the same as the de¬ 
crease in the other, which ex¬ 
plains why the sum of the 
tensions in motion is not quite 
the same as the sum of the ten¬ 
sions at rest. 

Again, when the belt, as it 
passes around the pulley, changes 
its motion from linear to circular, 
each particle of the belt—like a 
body whirling at the end of a cord 
about a center of rotation—tends, 
by centrifugal force, to fly away 
from the surface of the pulley, 
thereby decreasing the normal 
pressure, and hence the friction. 
The tensions in the belt between 
the pulleys are also changed some¬ 
what by centrifugal force, and as this increases as the square of the 
linear velocity, it is evident that the effect is greater at high speeds 
than at moderate or low speeds. 

A further circumstance that affects the driving power of a belt 
is the stiffness of the leather or other material of which it is made. 



The belt as it passes around the pulley assumes a circular form, and 
again straightens out as it leaves the pulley. Hence the theoretically 
perfect action is modified somewhat according to the sharpness of 
the bending and the thickness or flexibility of the belt; in other words, 














MACHINE DESIGN 


85 


a small pulley carrying a thick belt would be the worst case for suc¬ 
cessful calculation on a theoretical basis. 

Theory. The condition of the tight and loose sides of a belt 
transmitting power, is similar to that of the weighted strap and 
fixed pulley show T n in Fig. 50. If motion is desired of the strap 
around the pulley, it is necessary to make the weight W 2 of such a 
magnitude that it will overcome not only the weight W v but also 
the friction between the strap and the pulley. The strap tension 
T a is, of course, equal to W 2 and T 0 to W v The equation showing 
the balance of forces for the condition when motion is about to occur, is 

T a — T 0 = F = P (driving force) (39) 

If the pulley be free to turn on its axis, instead of being fixed 
as in Fig. 50, the strap by its friction on the pulley will turn the pulley, 
and the force of friction F becomes the driving force for the pulley 
as noted in equation 39. 

In Fig. 51, it may be supposed that W is a weight representing 



the resistance to be overcome. The tensions T a and T 0 , which are 
equal at first owing to stretching the belt tightly over the pulleys at 
rest, change when an attempt is made to raise the weight by turn¬ 
ing the larger pulley; and just as the weight leaves the floor, the 
equality of moments about the axis of the driven pulley gives the 
following equation: 

(T a — T 0 )r=FXr = PXr = WXr l (40) 

This equality of moments remains as long as the motion of the 
























86 


MACHINE DESIGN 


weight is uniform, and represents closely the conditions under which 
belt pulleys work. 

Although it may be determined from the above what the differ¬ 
ence of the belt tensions is, and what this difference will do when 
applied to the surface of a given pulley, the values of neither T a or 
T 0 are yet actually known, and until they are known the belt cannot 
be correctly proportioned. 

The tension on the tight side of the belt may be divided into 
three parts: tension doing useful work; tension from centrifugal 
force; and tension to keep belt from slipping. The tension upon 
the loose side may be divided into two parts: tension from centrifugal 
force and tension to keep belt from slipping. 

By means of higher mathematics a relation between these quan¬ 
tities may be determined which, when combined -with equations 
39 and 40, will allow the determination of the values of T a and T 0 . 

Practical Modification. In ordinary practice it has been found 
that a tension in the belt equal to the driving force will be sufficient 
to prevent slipping. 

The tighter the belt is drawn up, the greater is the pressure 
against the pulley, and hence the greater is the force of friction. 
But if the belt is pulled up too tightly, when driving is begun, the 
tension on the tight side becomes too great, and the belt breaks or 
is under such stress that it wears out quickly. Moreover, the great 
side pressure on the bearings carrying the shaft produces excessive 
friction, and the drive is inefficient. This is why a narrow belt driven 
at high speed is more efficient than a wide belt driven at slow speed, 
for although the former cannot be pulled up as tightly as the latter 
without overstraining it, yet by running it at high speed the required 
power may be obtained from the narrow belt. 

The centrifugal force is of small importance for low speeds, say 
of 3,000 feet per minute and less, and, therefore, it may usually be 
neglected. 

The angle of contact of belt with pulley is important, as a large 
value gives a great difference between the tension on the tight and 
loose sides; and it is desirable to make this difference as great as 
possible, because thereby the driving force is increased. The loose 
side of a horizontal belt should always.be above, as then the natural 
sag of the loose side due to its slackness tends to increase the angle 


MACHINE DESIGN 


87 


of contact with the pulley, while the tightening up of the lower side 
acts against the sag to make the loss of wrap as little as possible. 
Vertical belts which have the driving pulley uppermost, utilize the 
weight of the belt to increase the pressure against the surface of the 
pulley, slightly increasing its capacity for driving. The angle of 
contact may be artifically increased by a tightening pulley which 
presses the belt further around the pulley than it would naturally lie. 
This device adds, however, the friction of its own bearing, and im¬ 
pairs the efficiency of the drive. 

Since many of the factors upon which belt calculation depends 
must be assumed, the following empirical rules are generally used 
by American engineers, and if used with judgment give safe results. 



b X V 

1,000 



For a double belt, assuming double strength, this becomes 


h. 


b X V 
500 



With large pulleys and moderate velocities, this may hold good. 
With small pulleys and high velocities, however, the uncertain stresses 
induced by the bending of the fibers of the belt around the pulley, 
and the relatively great loss due to centrifugal force, modify this rela¬ 
tion, and a safer value for a double belt of the ordinary kind is 


or, still safer, 



b X V 
540 


h. 


b X v 





The theoretical value of horse-power for belt transmission is 

, _ PXV 

' P ' 33000 



Putting this value equal to the empirical value 

_PXV b XV 
' P ' 33000 1000 


and solving for P 


P = 336 









88 


MACHINE DESIGN 


This develops the fact that the empirical rule assumes a driv¬ 
ing force of 33 pounds per inch of width of single belt. 

Another way of expressing equation 42 is: A single belt will 
transmit one horse-power for every inch of width at a belt speed of 
1,000 feet per minute. 

Strength of Leather Belting. The breaking tensile strength 
of leather belting varies from 3,000 to 5,000 pounds per square inch. 
Joints are made by lacing, by metal fasteners, or by cementing. The 
strength of a laced joint may be about tV) of a metal-fastened joint 
about J, and of a cemented joint about equal to the full strength of 
the cross-sectional area of the belt. The proper working strength of 
belting depends on the use to which the belt is put. A continuously 
running belt should have a low tension in order to have long life and a 
minimum loss of time for repairs. For double leather belting it has 
been shown that a working tension of 240 pounds per square inch 
of sectional area gives an annual cost—for repairs, maintenance, and 
renewals—of 14 per cent of first cost. At 400 pounds working tension, 
the annual expense becomes 37 per cent of first cost. These results 
apply to belts running continuously; larger values may be used where 
the full load comes on but a short time, as in the case of dynamos. 

Good average values for working tensions of leather belts are: 

Cemented joints, 400 pounds per square inch 

Laced joints, 300 pounds per square inch 

Metal joints, 250 pounds per square inch 

It is evident from the equation for the theoretical value of horse¬ 
power that the horse-power of a belt depends upon two things, the 
driving force P and the velocity V, the driving force being the difference 
in tension on the tight and loose sides of the pulley. If either of these 
factors is increased, the horse-power is increased. Increasing P 
means a tight belt. Hence a tight belt and high speed together 
give maximum horse-power. But a tight belt means more side strain 
on shaft and journal. Therefore, from the standpoint of efficiency, 
a narrow belt under low tension at as high a speed as possible 
is used. 

Speed of Belting. The most economical driving speed for belts 
is somewhere between 4,000 and 5,000 feet per minute. Above these 
values the life of the belt is shortened; also “flapping,” “chasing,” 
and centrifugal force cause considerable loss of power. The limit of 


MACHINE DESIGN 


89 


speed with cast-iron pulleys is fixed at the same limit for bursting 
of the rim, which may be taken at one mile per minute. 

Material of Belting. Oak-tanned leather, made from the part 
of the hide which covers the back of the ox, gives the best results 
for leather belting. The thickness of the leather varies from .18 
to .25 inch. It weighs from .03 to .04 pound per cubic inch. The 
average thickness of double leather belts may be taken as .33 inch, 
although they may be ordered light, medium, or heavy, varying 
accordingly from J inch to T \ inch in thickness. Double leather 
belts are made by cementing the flesh sides of two thicknesses of 
belt together, leaving the grain or hair side exposed to surface wear. 
In a single-thickness belt, the grain side should be next to the pulley, 
as the flesh side is the stronger and is, therefore, better able to resist 
the tensile stress due to bending set up where the belt makes and 
leaves contact with the pulley face. 

Raw hide and semi-raw hide belts have a slightly higher co¬ 
efficient of friction than ordinary tanned belts. They are useful in 
damp places. The stress of these belts is about one and one-half 
times that of tanned leather. 

Cotton, cotton-leather, rubber, and leather link belting are 
some of the forms on the market, each of which is especially adapted 
to certain uses. For their weights and their tensile and working 
strengths consult the manufacturers’ catalogues. 

The practice of a prominent manufacturer in regard to the sizes 
of leather belting will be found useful for comparison, and is indicated 
in Table VII. 

Initial Tension in Belt. On the assumption that the sum of 
the tensions is unchanged, whether the belt be at rest or driving, 
there should be the following relation 

T a + T 0 = 2 T 


whence 



T a + T a 
2 



This is not strictly true, however, as is stated in the “Analysis of 
Belts.” It has been found that in a horizontal belt working at about 
400 pounds tension per square inch on the tight side, and having 
2 per cent slip on cast-iron pulleys, i. e., the surface of the driven 
pulley moving 2 per cent slower than that of the driver, the increase 



90 


MACHINE DESIGN 


TABLE VII 

Sizes of Leather Belting 


Width 

Thickness 

Width 

Thickness 

Single 

Double 

Single 

Double 

1 inch 

F2 i ncl1 

%, inch 

6 inch 

3*2 inch 

§ inch 

2 “ 

T 3 ? 

%o “ 

10 “ 

5 “ 

i “ 

3 “ 

A “ 

1 “ 

12 “ 


f “ 

4 “ 

h “ 

t “ 

14 “ 


13 “ 

32 

5 “ 

7 << 

3 2 

i “ 

20 “ 


_3L “ 

IF 


of the sum of the tensions when in motion over the sum of the ten¬ 
sions at rest, may be taken at about £ the value of the tensions at 
rest. Expressing this in the form of an equation 

T.+ T, = ±(2T) 

t = A(r n + t.) (47) 

The value of T thus found would be the pounds initial tension to 
which the belt should be pulled up when being laced, in order to 
produce T a and T 0 when driving. This value is not of very great 
practical importance, as the proper tightness of belt is usually secured 
by trial , by tightening pulleys , by pulley adjustment (as in motor 
drives), or by shortening the belt from time to time as needed. It is 
worth noting, however, that for the most economical life of the belt 
it would be very desirable in every case to weigh the tension by a spring 
balance when giving the belt its initial tension. This, however, 
is not always easy or even feasible; hence, it is a refinement with 
which good practice usually dispenses, except in the case of large and 
heavy belts. 

Care of Belts. Leather belts should be well protected from 
water, steam, or other moisture, and only the firmest, finest grained 
leather should be used in damp localities. Oil should never be 
allowed to drop on belts as it destroys the life of the leather. In 
order to keep a belt in good condition, warm tallow should be applied 
whenever the leather appears very dry. When belts are to be used 
in damp places the addition of a little resin to the tallow will pre¬ 
serve their strength. Too much resin, however, will leave the belt 
sticky and will cause cracking. 






















MACHINE DESIGN 


91 


Methods for Joining Ends of Belt. The ends of narrow, light 
belts such as are used on lathes, planers, and other machine tools, 
are generally fastened together by using strips of white leather tanned 
with alum. This form of joint is easily made, is flexible, and runs 
smoothly and noiselessly over the pulleys. Two common methods 
of lacing are shown in Figs. 52 and 53. To make a joint, first cut 
off the ends of the belt squarely. A punch should then be used for 
making the holes which must not come too near the ends or sides of 
the belt. Begin to lace in the middle of the belt and lace both sides 




Fig. 52. Simple Laced Joint 




Fig. 53. Strong Laced Joint 


with equal tightness. In Fig. 52, A is the outside and B the inside 
of the belt. Draw the lacing half way through the middle hole a 
from the under side, and then pass one end down through b, up through 
c, down through b and up through c again, then down through d to e 
where a cut is made upon the end of the lacing which acts as a barb 
to prevent unlacing. Lace the other side of the belt by carrying 
the other end up through d, down /, up g , down / and up g again, 
down a to h. 

A somewhat stronger form of lacing is shown in Fig. 53. Bring 
the lacing up through a, and continue through b, c, d, e, /, g, d, e, b, c, 
h, and i. The edges of the holes should not come nearer than § 
inch from the ends of the belts or f inch from the sides; and when 
lacing with two rows of holes, the second row should beat least If 
inches from the end. For 3-inch belts the number of holes may be 
the same as in Fig. 52. A 6-inch belt should have seven holes— 
four in the row nearest the end—and a 10-inch belt, nine holes. 
















































92 


MACHINE DESIGN 


For any style of lacing, care should always be taken that the 
lacing on the side of the belt next the pulleys is parallel with the 
edge of the belt. When heavy belts transmit large amounts of power, 
their joints are generally made by gluing and riveting. 

Example. The tension on the tight side of a belt is 1,059 
pounds, and on the loose side 645 pounds; the pulley is 42 inches in 
diameter and the speed is 470 revolutions per minute. What is the 
horse-power transmitted and what should the width be for a double 
belt? 

Solution. 

P = 1059 — 645 = 414 pounds 

42 

V = 470 X 3.1416 X —- =5168 ft. per minute 


, 414X5168 . 

h • P- = "33ooo~ = -65 h ‘ p -> nearly 

Above 3,000 feet per minute, centrifugal force should be considered 
and, therefore, in order to obtain the width of the double belt, use 

b X V 


the formula, h. p. = 


b = 


700 
64 = 

64 X 700 
5168 


Then 

b X 5168 
700 

= 8f" nearly. Ans. 


PROBLEMS FOR PRACTICE 

1. A double belt transmits 50 horse-power at 2,000 feet per 
minute. What should be the width? 

2. A pulley 24 inches in diameter revolving at 250 revolutions 
per minute is to transmit 25 horse-power. Determine the width 
of belt. 

3. A belt with a velocity of 3,500 feet per minute has a tension 
of 969 pounds in its tight side and 497 pounds in its loose side. 
What horse-power is it transmitting? 








































































































































































































MACHINE DESIGN 

PART II 


ROPE GEARING 


NOTATION—The following notation is used throughout the chapter on Rope 

Gearing: 


A = Net area of rope (sq. in.) 
a = Diameter of wire in a rope 
(inches) 

B = Total stress due to bending (lbs.) 
C = Centrifugal force (lbs.) 

D = Distance between sheaves (feet) 
d = Diameter of the rope 
E = Coefficient of elasticity 
H = Horse-power 

L = Distance between pulleys (feet) 


P = Coefficient of friction 
P = Driving force at pulley rim (lbs.) 
R = Radius of pulley (inches) 

S = Sag of rope (feet) 

T = Tension in rope (lbs.) 

T a = Total tension on tight side (lbs.) 
T 0 = Total tension on slack side (lbs.) 
t = Tension, balancing slipping (lbs.) 
V = Velocity of rope (ft. per sec.) 
W — Weight of one foot of rope (lbs.) 


HEMP AND COTTON ROPB 


Analysis. The stresses acting upon a rope drive are practically 
the same as those acting upon a belt and the analysis for belts will 
apply to ropes. 

Theory. The equation showing the balance between the actual 
working force and the tension in the rope is 

P=T a -T 0 (48) 

The formula for centrifugal force in belts is 


C = 


JVV 2 

32 



The balance of the forces acting on the loose side is expressed 
by the equation 

T 0 = C 4- t (50) 

Since one horse-power equals 550 foot-pounds per second 


H 


P X V 



4 - 

»• 


550 





94 


MACHINE DESIGN 




The rims of pulleys for hemp or cotton-rope gearing are usually 
grooved as shown in Fig. 54. In a semicircular groove, Fig. 55, 


Fig. 54. Wedge 
Groove 


Fig. 55. Semicir¬ 
cular Groove 


Fig. 56. Flat Run 


the friction is very little more than that upon an ordinary flat cylin¬ 
drical pulley of the shape shown in Fig. 56. The frictional resistance 
of the wedge-shaped groove may be determined from a parallelogram 
of forces. Referring to Fig. 56, it is evident that the pressure be¬ 
tween the rope and pulley is equal to the radial force produced by 
the tension in the rope. In the wedge-shaped groove the radial force 
has been resolved into two components as shown in Fig. 57. Draw¬ 
ing a parallelogram of forces, Fig. 58 is 
obtained. If the rope slipped it would be 
compelled to slip on both sides, then the 
total resistance to slipping would be 
fi(K l + K 2 ). It will be noted that K\ + K 2 
will always be greater than Q unless a is 
equal to 180° and the less angle a is, the 
greater will be K l + I\ 2 relative to Q. 

Practical Modification. It is usual in 
practice to make the sides of the groove 
incline at an angle of 45° to each other. 
Under these conditions experience has proved that the tension to 
prevent slipping will be sufficient when it is equal to one-half the 

P 

force doing useful work, or t = —. 

Transmission ropes wear internally, due to the fibers sliding 
on one another when the rope bends about the pulley or sheaves. 
In order to decrease this wear as much as possible it has been found 


Fig. 57. Components of Forces 





















































































I 


MACHINE DESIGN 


95 


advisable to use a sheave whose diameter is at least forty times that 
of the rope. The use of a large sheave not only adds to the life of 
the rope but offers a larger surface of contact and increases the driv¬ 
ing force. 

Wooden sheaves are not suitable for rope transmission because 
it is impossible to obtain such a sheave that will be of uniform density 
throughout, thus causing more wear in some portions of the groove 
than in others. This results in uneven running and increased wear 
on the rope. 

Materials. Cotton rope is generally used on small machines 
as it is soft and flexible. Hemp rope possesses great strength, is 
durable, and withstands exposures to the 
weather. Ropes are usually made of three, 
four, or six strands. Their diameter is 
from 1 to If inches; however, the size is 
sometimes given by the girth or circum¬ 
ference. The net section is about .9 of 
the total area of the rope. 

It has been shown by experiment that the weight of dry rope 
may be determined by the formula 

W = .3 d 2 



Fig. 58. Resolution of Force 


From the best information obtainable the breaking strength may be 
taken as 7,000 lbs. per sq. in., and the factor of safety 35 making 
the working strength 200 lbs. per sq. in. 

It is not advisable to run ropes at a speed greater than 5,000 
feet per minute as the centrifugal force begins to greatly reduce the 
friction between the rope and the pulley. 

The sag on the slack side varies with the speed and the power 
transmitted. When at rest the sag is not as great as when running, 
being greater on the tight side or less on the slack side. The sag 
on the driving side when transmitting normal horse-power is the same, 
no matter what size of rope is used or what its velocity may be. Ex¬ 
perience has shown that the sag may be determined by the formula 



WD 2 
8 T a 



To determine the sag on 


the slack side use the formula 



WD 2 

8 T 0 


(53) 





96 


MACHINE DESIGN 


I 


Systems of Rope Driving. There are two systems of rope driving 
in general use, one known as the multiple, or English , and the other 
as the continuous, or American. 

The multiple system consists of one or more independent ropes 
running side by side in the grooves of the sheaves. This system is 
suitable for the transmission of large power and gives best results 
where the shafts are parallel or nearly so and where the drive is 
sufficiently off the vertical to prevent the ropes when slack from 
leaving the pulley. If one rope breaks this does not cause stoppage 
of the entire drive. 

The continuous system makes use of one rope which is wound 
around the driving and driven pulley several times. In this system, 
Fig. 59, the rope is conducted from the outside groove of one pulley 


oWVEF? 



Fig. 59. Continuous Rope Drive 


to the inside groove of the other. This is accomplished by using 
a traveling tension carriage or jockey. The carriage also serves 
to maintain a uniform tension throughout the rope and is arranged 
to travel back and forth, automatically regulating the slack; thus, 
the stretch in the rope and the inequalities in the load are taken care of. 

A system patented by Joseph H. Hoadley uses a winding or 
idler pulley. The smaller pulley is provided with more grooves 
than the larger. After the rope has filled all the grooves of the 
larger and an equal number of grooves of the smaller, the rope is 
then carried around the grooves of the winding pulley. In this 
way the contact surface of the smaller pulley is increased and made 
more nearly equal to the larger. 
























MACHINE DESIGN 


97 


WIRE ROPE 


Theory. The forces acting in wire rope transmission are of the 
same character as those acting in belts. In addition, the stress due 
to the rope bending about the pulley must be considered and is 
represented by the equation 


B = 


AEa 
2 R 



I he total stress due to centrifugal force is as before 


C = 


WV 2 

32 


The equation for the driving force is 


T - T 

n -*-o 


The horse-power transmitted is 


H = 


P X V 
550 


The deflection of the rope may be calculated by means of the following 
formula—which will give results close enough for practice— 



If it is wished to know the tension on the rope, the formula may be 
changed to read 



WL 2 

~8S~ 


+ TVS 



Practical Modification. Wire rope is especially adapted to the 
transmission of large power over great distances. It is unsuitable 
for small power and short drives because of its rigidity and its short 
life when bent about pulleys of small diameter. 

It has been found that with wire rope under best conditions the 

? 

tension required to prevent slipping is equal to about—. 

The force at which the rope is worked should not exceed the 
elastic limit which may be taken as 57,000 lbs. per square inch for 
steel and 28,500 lbs. per square inch for wrought iron. For the 
greatest safe working tension there is a certain ratio existing between 
the diameter of the sheave and the diameter of the wires composing 
the rope. The minimum diameters of sheaves for various ropes is 











98 


MACHINE DESIGN 


usually given in rope manufacturer’s catalogues. The coefficient of 
elasticity may be taken as 28,000,000 for both iron and steel. 

Pulleys for Wire Ropes. Wire ropes are injured by the lateral 
crushing if run upon pulleys with V-shaped grooves. Hence, pulleys 
are constructed with wide grooves so that the rope may rest upon 

the bottom. The frictional resistance between 
the rope and the pulley is much increased by 
lining the bottom of the groove with gutta-percha, 
tarred hemp, rubber, leather, or wood. The 
material used is hammered into the groove, often 
dovetailed in section. Where the working tension 
is very great, wood filling is preferred. A cross- 
section of a wire rope sheave is shown in Fig. 60. 

Fig. 60 . Groove with The weight per foot of wire rope equals 1.43d 2 . 

Wood piiimg Examples. 1. A drive pulley 72 inches in di¬ 
ameter revolving at a speed of 100 revolutions per minute and having 
four grooves using l^-inch hemp rope will transmit what horse-power? 

Solution. The problem may be solved by first finding the 
horse-power transmitted by one rope and then multiplying by 4 to 
find the total. 

W = .3d 2 = .3 X 1.5 2 = .675 lb. per foot 
Maximum tension allowed in the rope is 

.7854 X 1.5 2 X .9 X 200 = 318.1 lbs. 



72 


100 


V = 3.1416 X-^X = 31.4 feet per second 
C 


12 60 
■67S X 3L 4 2 = 20 7 , b 
32 

P 


2 


+ C 


P = T n - T, = T. - + C ) 


fP-T m -C 

P=^-(T.-C) = -1 (318.1 - 20.7) = 

tr _ 198.2 X 31.4 , 

550 

Total horse-power = 11.3 X 4 = 45 nearly. 


198.2 lbs. 















MACHINE DESIGN 


99 


2 . Power is transmitted by a |-inch 6-strand 19-wire steel 
rope. The driving pulley is 8 feet in diameter, the driven pulley 
2 feet, and the speed of the rope 3,600 feet per minute. The net 
area of the wire is 0.097504 square inch, and the diameter of the 
wires composing the rope, 0.033 inch. The strength of wire is 
57,000 pounds per square inch. Find the horsepower which may 
be transmitted. 


Solution. 

W 


1.43 X .5 2 = .3575 lb. per foot 


V = = 60 feet per second 

The tension due to bending is greater at the small pulley. 

B = .097504 X 2800 0 ° 00 X - 033 = 3754 ib. 

2 X 12 

_ WV* .3575 X 60* „ ,, 

C ~ ~32 32 4tU lb ‘ 

The tension on the slack side would be the sum of the ten¬ 
sions due to bending, centrifugal force, and that necessary to keep 
the rope from slipping, or 

T 0 = ~+B+C = ~ + 3754 + 40.2 = -j + 3794.2 

P=T a -T a 

P = T u - (-£- + 3794.2) = T n -~- 3794.2 

2 

= T q - 3794.2 P = ( T n - 3794.2) 

2 o 

T u = 57000 X .097504 = 5557.7 lb. 
p = A ( 5557.7 - 3794.2) = ~ X 1763.5 = 1175.67 

O O 

„ 1175.67 X 60 _ , OQ 

550 

What is the sag in the tight side of the wire if the distance 
between pulleys is 100 feet? 

5557. 


8 = 


2 X .3575 


■2--J 

1575 V 


5557.7 2 


100 2 


4 X .3575 2 8 


= 7773 — 7772.92 = 0.08 feet = 1 inch nearly 

















100 


MACHINE DESIGN 


PROBLEMS FOR PRACTICE 

1 . Find the horse-power which may be transmitted by a J-inch 
manila rope running at a speed of 3,000 feet per minute. 

2 . In the preceding problem if the distance between pulleys is 
25 feet, what is the sag on both the tight and loose sides? 

3. Will a J-inch 7-wire steel rope composed of "wires .028 inch 
in diameter and having a net area of .025862 sq. in. transmit 45 
horse-power? The velocity of the rope is 2,400 feet per minute and 
the small pulley is 24 inches in diameter. 

4. A J-inch 19-wire iron rope transmits 150 horse-power. 
The small pulley is the driver, 5 feet in diameter, and operates 
at a speed of 250 revolutions per minute. The rope is composed of 
wires .033 inch in diameter and has a net area of .097504 sq. in. 
Is this iron rope sufficiently large to withstand the strain? 

PULLEYS 

NOTATION—The following notation is used throughout the chapter on 

Pulleys: 

l = Length of hub (inches) 
iY = Number of arms 
n = Number of rim bolts, each side 
P = Driving force of belt (lbs.) 

Pi = Force at circum. of shaft (lbs.) 

P 2 = Force at circum. of hub (lbs.) 
p = Stress in rim due to centrifugal 
force (lbs. per sq. in.) 

R = Radius of pulley (inches) 

S = Fiber stress (lbs. per sq. in.) 

N s = Shearing stress (lbs. per sq. in.) 

T = Thickness of web (inches) 
t = Thickness of rim (inches) 
t 2 = Thickness of rim bolt flange 
(inches) 

v = Velocity of rim (ft. per sec.) 
w — Weight of material (lbs. per cu. 
in.) 

Analysis. If a flexible band be wrapped completely about 
a pulley, and a heavy stress be put upon each end of the band, the 


A = Area of rim (sq. in.) 
a — Area of arm (sq. in.) 
b = Center of pulley to center of belt 
(inches; practically equal to R) 
C 1 = Total centrifugal force of rim 
(lbs.) 

c = Distance from neutral axis to 
outer fiber (inches) 

D = Diameter of pulley (inches) 

D x = Diameter of hub (inches) 
d x — Diameter of bolt at root of 
thread (inches) 

d = Diameter of bolt holes (inches) 
g = Acceleration due to gravity (ft. 
per sec.) 

h = Width of arm at any section 
(inches) 

I = Moment of inertia 
L = Length of arm, center of belt to 
hub (inches) 

Lj = Length of rim flange of split 
pulley (inches) 


MACHINE DESIGN 


101 


rim of the pulley will tend to collapse just like a boiler tube with 
steam pressure on the outside of it. A compressive stress is in¬ 
duced which is very nearly evenly distributed over the cross-section 
of the rim, except at points where the arms are connected thereto. 
At these points the arms, acting like rigid posts, take this compres¬ 
sive stress. Now, a pulley never has a belt wrapped completely 
round it, the fraction of the circumference embraced by the belt 
being usually about \, and seldom reaching j, even with a tightener 
pulley. Assuming the wrap to be J the circumference, and that 
all the side pull of the belt comes on the rim, none being transmitted 
through the arms to the hub, then one half of the rim is pressed hard 
against the other half by a force equal to the resultant of the belt 
tensions, which, in this case, would be the sum of them. Dividing 
the pulley by a plane through its center and perpendicular to the 
belt, the cross-section of the pulley rim cut by this plane has to take 
this compressive stress. 

This analysis is satisfactory from an ideal standpoint only, for 
the intensity of stress due to the direct pull of the belt, with the usual 
practical proportions of rim, would be very small. Moreover, the 
element of speed has not been considered. 

When the pulley is under speed, a set of conditions which com¬ 
plicates matters is introduced. The centrifugal force due to the 
weight of the rim and arms is no longer negligible, but has an im¬ 
portant influence upon the design and material used. This cen¬ 
trifugal force acts against the effect of the belt-wrap, and tends to 
reduce the compressive stress, or, overcoming the latter entirely, 
sets up a tensional stress both in the rim and in the arms. It also 
tends to distort the rim from a true circle by bowing it out between 
the arms, thus producing a bending moment in the rim, which is at 
a maximum at the points where the rim joins each arm. 

It can readily be imagined that the analysis in detail of these 
various stresses in the rim acting in conjunction with each other 
is quite complicated—far too much so in fact, to be introduced here. 
As in most cases of such design, however, one controlling influence 
can be separated out from the others, and the design based thereon 
with sufficient margin of strength to satisfy the more obscure con¬ 
ditions. This is rational treatment, and the “theory” will be studied 
accordingly. 


102 


MACHINE DESIGN 


The rim, being fastened to the ends of the arms, tends, when 
driving, to be sheared off, and the resistance to shear depends upon 
the areas of the cross-sections of the arms at their point of joining 
the rim. The force that produces this shearing tendency is the 
driving force of the belt, or the difference between the tensions of the 
tight and loose sides. 

Again, at the point of connection of the arms to the hub, a shear¬ 
ing action takes place, so that, if this shearing tendency w T ere carried 
to rupture, the hub would literally be torn out of the arms. Now, 
viewing the arms as beams loaded at the end with the driving force 
of the belt, and fixed at the hub, a heavy bending stress is set up, 
which is maximum at the point of connection to the hub. If the 
rim were stiff enough to distribute this driving force equally between 
the arms, each arm would take its proportional share of the load. 
The rim, however, is quite thin and flexible; and it is not safe to 
assume this perfect distribution. It is usual to consider that one half 
the whole number of arms take the full driving force. 

Theory. Pulley Rim. Evidently it is practically impossible 
to make so thin a rim that it will collapse under the pull of a belt. 
As far as the theory of the rim is concerned, its proportion probably 
depends more upon the calculation for centrifugal force than upon 
anything else. 

In order to separate this action from that of any other forces, 
let it be supposed that the rim is entirely free from the arms and hub, 
and is rotating about its center. Every particle, by centrifugal 
force, tends to fly radially outward from the center. The tendency 
with which one half of the rim tends to fly apart from the other 
half is indicated by the force C x ; and the relation between C t and 
the small radial force c for each unit-length of rim can readily be 
found from the principles of mechanics. The case is exactly like 
that of a boiler or a thin pipe subjected to uniform internal pres¬ 
sure, which, if carried to rupture, would split the rim along a longi¬ 
tudinal seam. 

The tensile stress thus induced per square inch may be found, 
by simple mechanics, to be 

12 wv 2 
p ~ 

or, since w = 0.26 pound for cast iron, and g = 32.2 feet per second. 




MACHINE DESIGN 


103 


V 


0.097 v 5 


(say, 




and, if p be taken equal to 1,000 pounds per square inch, which is 
as high as it is safe to work cast iron in this place, 


v = 100 feet per second 



This shows the curious fact that the intensity of stress in the rim 
is directly proportional to the square of the linear velocity, and 
wholly independent of the area of cross-section. It is also to be 
noted that 100 feet per second is about the limit of speed for cast- 
iron pulleys to be safe against bursting. 

By theoretically considering the rim together with the arms as 
actually connected to it, a much more complicated relation is ob¬ 
tained. This condition causes the rim to expand more than the 
arms and to bulge out between them. This makes the rim act some¬ 
thing like a continuous beam uniformly loaded; but even then the 
resulting stress is not clearly defined on account of the variable stretch 
in the arms. Investigation on this basis is not needed further than 
to note that it is theoretically better, in the case of a split pulley, to 
make the joint close to the arms, rather than in the middle of a span. 

Pulley Arms. The centrifugal force developed by the rim 
and arms tends to pull the arms from the huh. On the belt side, 
this is balanced to some extent by the belt wrap', which tends to 
compress the arm and to relieve the tension. On the side away 
from the belt, the centrifugal action has full play, but the arm is 
usually of such cross-section that the intensity of this stress is very 
low and may safely be neglected. 

The rim being very thin in most cases, its distributing effect 
cannot be depended on, hence the driving force of the belt may be 
taken entirely by the arms immediately under the portion of the belt 
in contact with the pulley face. For a wrap of 180° this means that 
only one half of the pulley arms can be considered as effective in 
transmitting the turning effort to the hub. Each of these arms is a 
lever fixed at one end to the hub and loaded at the other. A lever 
of this description is called a cantilever beam, its maximum moment 
existing at its fixed end. The load that each of these beams may be 


subjected to is-^, and, therefore, 


the maximum external moment at the 



104 


MACHINE DESIGN 


9 pj 

hub is-. From mechanics, the internal moment of resistance 

N 

SI 

of any beam section is —, and equilibrium of the beam can be 

satisfied only when the external moment is equal to the internal moment 
of resistance of the beam section. Equating these two 


2 PL 
N 


SJ 

c 


(60) 



The arms of a pulley are usually of the 
elliptical or segmental cross-section, and may 
be of the proportions shown in Fig. 61. For 

either of these sections the fraction — 


is 


Fig. 61. Elliptical Arms 


approximately equal to 0.0393 h 3 . For con¬ 
venience—the error caused being on the 
safe side —L may be taken as equal to the 
full radius of the pulley R , whence 

2 PR -(61) 


N 


= 0.0393S/t 3 


in which S may be from 2,000 to 2,250 for cast iron. 

Taking moments about the center of the pulley, and solving for 
P 2 , the force acting at the circumference of the hub. 


2 PR 
N 


P-A 


or 


p = A 13 

2 ND 


(62) 


The area of an elliptical section is n times the product of t 'e 
half axes. With the proportions of Fig. 61 this becomes 

71/(2 (63) 


a = it X 0.2& X 0.5 h 


10 


Equating the external force to the internal shearing resistance 

4 PR 7ih 2 S . 


ND, 


10 


or 


























MACHINE DESIGN 


105 


h= I 40 PR 

N D.NttS, 


(64) 


in which the shearing stress S B may run from 1,500 to 1,800 for cast 
iron. 

Although both bending and shearing stresses as calculated 
above exist at the base of the arms, the bending is, in practically 
every case, the controlling factor in the design of the arms. An 
arm-section large enough to resist bending would have a very low 
intensity of shear. 

If the number of arms be increased indefinitely, a continuous 
arm or web is finally reached, in which the bending action is elimi¬ 
nated. It may still shear off at the hub, where the area of metal 
is the least— i. e., at minimum circumference. In this case the area 
under shearing stress is nD^T; and the force at the circumference 
of the hub is 


PR 2 PR 



Equating external force to internal shearing resistance 


2 PR 

D i 


nD 1 TS, 



ar 



2 PR 
tiD 2 T 


(65) 


Pulley Hub . As in the case of the arms, centrifugal force does 
not play much part in the design of the hub of a pulley. The hub 
is designed principally to carry the key, and through it transmit the 
turning moment to the shaft. Considered thus, the hub may tear 
along the line of the key or crush in front of the key. 

For example, in Fig. 62, if the connection with the lower arms 
be neglected, and the upper arms be held fast while a turning force 
P v at the surface of the shaft, is transmitted to the hub through the 
key, then the metal of the hub directly in front of the key is under 
crushing stress; and the metal along the line eb, from the corner to 
the outside, is under tensile stress. This condition is the worst that 
could possibly happen, because the bracing effect of the lower arms 
has been neglected, and the key is located between the arms. 









106 


MACHINE DESIGN 


Taking moments about the center of the shaft, the value of the 
force at the shaft circumference, or the “key pull,” is 

Pr = ^ (66) 

P k 

Now —- = —, k being the distance from the center of the shaft 
r 3 r 

to the center of eb, and the area of metal which is subjected to the 
tearing action P 3 is / X eb. Equating the external force to the 
internal resistance, and assuming that the stress is equally distributed 
over the area l X eb 


or 



P P 

—— = S X l X eb 
r 



PR 

k X l X eb 



The intensity of crushing on the metal in front of the key, due 
to force P v depends upon the thickness of the key, and is properly 
discussed later under “Keys.” 

Practical Modification. Pulley Rim. The theoretical cal¬ 
culation for the thickness of the rim may give a thickness that could 

not be cast in the foundry, and 
the section in that case will 
have to be increased. As light 
a section as can be readily cast 
will usually be found abun¬ 
dantly strong for the forces it 
has to resist. A minimum 
thickness at the edge of the 
rim is about inch; and as 
the pulleys increase in size, 
the rim also must be made 

Fig. 62. Distribution of Stresses thicker; otherwise the rim will 

cool so much more quickly 
than the arms, that the latter, on cooling, will develop shrinkage 
cracks at the point of junction. 

For a velocity of 6,000 feet per minute, it is found that the tension 
in pounds per square inch, in the rim, due to centrifugal force, is 
























MACHINE DESIGN 


107 


970. Though this in itself is a low value, yet the uncertain nature 
of cast iron, its condition of internal stress, due to casting, and the 
likely existence of hidden flaws and pockets, have established the 
usage of this figure as the highest safe limit for the peripheral speed of 
cast-iron pulleys. It is easily remembered that cast-iro 7 i 'pulleys 
are safe for a linear velocity of about one mile per minute. 

To prevent the belt from running off the pulley, a “crown” 
or rounding surface is given the rim; however, a tapered face, which 
is more easily produced in the ordinary shop, may be used. This 
taper should be as little as possible, consistent with the belt staying 
on the pulley; ^ inch per foot each way from the center is not too 
much for faces 4 inches wide and less; while above this width, \ inch 






r 


"T 

4 > 


////////« 



Fig. 63. Pulley Rims 


per foot is enough. As little as J inch total crown has been found 
to be sufficient on a 24-inch face, but this is probably too little for 
general service. Instead of being “crowned,” the pulley may be 
flanged at the edges; but flanged pulley rims chafe and wear the edge 
of the belt. 

The inside of the rim of a cast-iron pulley should have a taper 
of \ inch per foot to permit easy withdrawal from the foundry mould. 
This is known as draft. If the pattern be of metal, or if the pulley 
be machine-moulded, the greater truth of the casting does not re¬ 
quire that the inside of the rim be turned, as the pulley, at low speeds, 
will be in sufficiently good balance to run smoothly. For roughly 
moulded pulleys, and for use at high speed, however, it is necessary 
that the rim be turned on the inside to give the pulley a running 
balance. 

Fig. 63 shows a plain rim a, also one stiffened by a rib b. Where 
heavy arms are used this rib is essential so that there will not be 
too sudden change of section at the junction of rim and arm, and 
consequent cracks or spongy metal. 























108 


MACHINE DESIGN 


The following empirical formula is often used to determine the 
width of the rim 

B = + -4) (68) 

Split Pulleys. The split pulley is made in halves and provided 
with bolts through flanges and bosses on the hub for holding the two 
halves together. When the pulley is in place on the shaft, bolted 
up as one piece, it is subjected to the same forces as the simple pulley. 
Hence, its general design follows the same principles, and the fasten¬ 
ing of the two halves, and the effect of this fastening on the detail 
of rim and hub, only need be studied. Practical considerations are 
chiefly responsible for the location of the joint in a split pulley be¬ 
tween the arms instead of directly at the end of an arm, where 
theoretically it would seem to be required. It is usually more con¬ 
venient in, the foundry machine shop to have the joint between the 
arms; so it is generally placed there, and strength provided to per¬ 
mit this. It is possible, however, to provide a double arm, or a 
single split arm, in which case the joint of the pulley comes at the 
arm, and the “heeling” action of the rim flanges is prevented. 

The rim bolts should be. crowded as close as possible to the rim 
in order to reduce the stress on them, and also to reduce the stress 
in the flange itself. This practical point must not be forgotten, 
however, that the bolts must have sufficient clearance to be put into 
place beneath the rim. 

While it is evident that the rim bolts are most effective in taking 
care of the centrifugal action on the halves of the pulley, yet in small 
split pulleys it is quite common to omit the rim bolts and to use the 
hub bolts for the double purpose of clamping the shaft and holding 
the two halves together. The pulley is cast with its rim continuous 
throughout the full circle, and it is machined in this form. It is 
then cracked in two by a well-directed blow of a cold chisel, the 
casting being especially arranged for this along the division line by 
cores so set that but a narrow fin of metal holds the two parts together. 
This provides sufficient strength for casting and turning, but permits 
the cold chisel to break the connection easily. 

Pulley Arms . The arms should be well filleted at both rim 
and hub, to render the flow of metal free and uniform in the mould. 


MACHINE DESIGN 


109 


The general proportions of arms and connections to both hub and 
rim may perhaps be best developed by trial to scale on the draw¬ 
ing board. The base of the arm being determined, it may gradu¬ 
ally taper to the rim, where it takes about the relation of f to J the 
dimensions chosen at the hub. The taper may be modified until 
it looks right, and then the sizes checked for strength. 

Six arms are used in the great majority of pulleys. This num¬ 
ber not only looks well, but is adapted to the standard three-jawed 
chucks and common clamping devices found in most shops. Elliptical 
arms look better than the segmental style; the flat, rectangular arm, 
which gives a very clumsy and heavy appearance, is seldom found 
except on the very cheapest work. 

A double set of arms may be used on an excessively wide face, 
but it complicates the casting to some extent. 

Although a web pulley may be calculated for shear at the hub, 
yet a thickness of web intermediate between the thickness of the rim 
and that of the hub—which will satisfy the casting requirements— 
will meet the requirements as to strength. 

Pulley Hub. The hub should have a taper of \ inch per foot 
draft, similar to that of the inside of the rim. The length of the 
hub is arbitrary, but if made about f the face width of the pulley 
will prevent rocking on the shaft. 

The diameter of the hub, aside from the theoretical consideration 
given above, must be sufficient to take the wedging action of a taper 
key without splitting. This relation cannot well be calculated but 
probably the best rule that exists is the familiar one that the hub 
should be twice the diameter of the shaft , a rule which, however, can¬ 
not be literally adhered to, as it gives too small hubs for small shafts 
and too large ones for large shafts. It is always well to locate the 
key, if possible, underneath an arm instead of between the arms, 
thus gaining the additional strength due to the backing of the arm. 

Special Forms of Pulleys. The plain, cast-iron pulley has been 
used in the foregoing discussion as a basis of design. A pulley is, 
however, such a common commercial article, and finds such univer¬ 
sal use, that special forms, which can be bought in the open market, 
are not only cheaper but better than the plain, cast-iron pulley, at 
least, for regular line-shaft work. 

Cast iron is a treacnerous and uncertain material for rims of 


110 


MACHINE DESIGN 


pulleys. It is not well suited to high fiber stresses; hence the range 
of speed permissible for pulley rims of cast iron is limited. Steel 
and wrought iron, having several times the tensional strength of 
cast iron, and being, moreover, much more nearly homogeneous 
in texture, are well suited for this work; one of the best pulleys on 
the market consists of a steel rim riveted to a cast-iron spider. Such 
an arrangement combines strength and lightness, without increas¬ 
ing complication or expense. 

The all-steel pulley is a step further in this direction. Here 
the rim, arms, and hub are each pressed into shape by specially 
devised machinery, then riveted and bolted together. This pulley 
is strictly a manufactured article, which could not compete with the 
simpler form unless built in large quantities, enabling automatic 
machinery to be used. Large numbers of pulleys are built in this 
way, and are put on the market at reasonable prices. 

Wood-rim pulleys have been made for many years, and, ex¬ 
cept for their clumsy appearance, are excellent in many respects. 
The rim is built up of segments in much the same way as an ordi¬ 
nary pattern is made, the segments being so arranged that they 
will not shrink or twist out of shape from moisture. The hubs 
may be of cast iron, bolted to wooden webs, and carrying hard¬ 
wood split bushings, which may be varied in bore within certain 
limits so as to fit different sizes of shafting. The wooden pulley 
is readily and most often used in the split form, thus enabling it 
to be put in position easily at any point of a crowded shaft. It is 
often merely clamped in place, thus avoiding the use of keys or 
set screws, and not burring or roughening the shaft in any way. 

Example. The driving force of a belt on a 36-inch pulley is 
800 lbs. and the belt wrap, about 180°. Calculate proportions of the 
six elliptical arms to resist bending, the allowable fiber stress being 
2,000 lbs. per sq. in. 

Solution. Using equation (61) 

2 PR 

= .0393 Sh 3 
N 

S = 2000 R = 18 N = 6 P = 800 


and substituting the above values. 


2 X 800 X 18 


- .0393 X 2000/i 3 


C 




MACHINE DESIGN 


111 


h t = 2 X 800 X 18 = 

6 X .0393 X 2000 

h =iX61 = 3.9365= 4 inches, nearly. 

Then the thickness of the arm = .4X4= 1.6 or about If". 
Therefore, the dimensions of the arm should be 4" X If". 

PROBLEMS FOR PRACTICE 

1. Calculate the tensile stress due to centrifugal force in the 
rim of a 30-inch cast-iron pulley at 500 revolutions per minute. 

2. A pulley 12 inches in diameter, f-inch web, 4-inch diameter 
hub, transmits 25 horse-power at a belt speed of 3,000 ft. per minute. 
Calculate the maximum shearing stress in the web. 


SHAFTS 


NOTATION—The following notation is used throughout the chapter on 

Shafts: 


A°= Angular deflection (degrees) 

B = Distance between bearings (feet) 
c = Distance from neutral axis to 
outer fiber (inches) 
d, d 0 , d 2 , d 3 , d 4 = Diameters of shaft 
(inches) 

d, = Internal diameter of shaft 
(inches) 

E = Direct modulus of elasticity (a 
ratio) 

e = Transverse deflection (inches) 
G — Transverse modulus of elasticity 
(a ratio) 

H = Horse-power (33,000 ft.-lbs. per 
minute) 

I = Moment of inertia 
K = Distance between bearings 
(inches) 

L = Length along shaft (inches) 

Lj, Z/ a = Length of bearings (inches) 


M = Simple bending moment (inch- 
lbs.) 

M e — Equivalent bending moment 
inch-lbs.) 

N = Number of revolutions per min¬ 
ute 

P = Driving force of belt (lbs.) 

R = Radius at which load as stated 
acts (inches) 

S = Fiber stress, tension, compres¬ 
sion, or shearing (lbs. per 
sq. in.) 

T = Simple twisting moment (inch- 
lbs.) 

T % = Equivalent twisting moment 
(inch-lbs.) 

T n = Tension in tight side of belt (lbs.) 

T 0 = Tension in loose side of belt (lbs.) 

W = Load applied as stated (lbs.) 


Analysis. The simplest case of shaft loading is shown in Fig. 
64. The equal forces W, similarly applied to the disk at the dis¬ 
tance R from its center, tend to twist the shaft off, the tendency 
being equal at all points of the length L between the disk and the 



. 112 


MACHINE DESIGN 


post to which the shaft is rigidly fastened. The fastening to the 
post, of course, in this ideal case, takes the place of a resisting mem¬ 
ber of a machine. A state of pure torsion is induced in the shaft; 
and any element, such as ca, is distorted to the position cb, aob being 
the angular deflection for the distance L. 

The case, Fig. 65, is illustrative of what occurs when a belt 

pulley is substituted for the simple 
disk. Here the twisting action 
is caused by the driving force 
of the belt, which is T a — T Q =P, 
acting at the radius R. Torsion 
and angular deflection exist in the 
shaft, as in Fig. 64. In addition, 
however, another stress of a dif¬ 
ferent kind has been introduced; 
for not only does the shaft tend 
to be twisted off, but the forces 
T a and T 0 , acting together, tend 
to bend the shaft, the bending 
moment varying with every sec¬ 
tion of the shaft, being nothing 
at the point o, and maximum at 
the point c. This combined 
action is the most common of any that is found in ordinary machinery. 

In Fig. 65, if the forces T n and T 0 be made equal, there will be 
no tendency at all to twist off the shaft, but the bending will remain, 
being maximum at the point c. This condition is illustrative of the 
case of all ordinary pins and studs in machines. In this sense, a 
pin or a stud is simply a shaft which is fixed to the frame of the ma¬ 
chine, there being no tendency to turning of the pin or stud itself. 
The same condition would be realized if the disk in Fig. 65 were 
loose upon the shaft. In that case, the bending moment would be 
caused by T n + T 0 acting with the leverage L. Of course there 
would have to be some resistance for T a — T Q to work against, in 
order that torsion should not be transmitted through the shaft. 
This condition might be introduced by having a similar disk lock 
with the first one by means of lugs on its face, thus receiving and 
transmitting the torsion. 































MACHINE DESIGN 


113 


If the distance L becomes very great, both the angular deflec¬ 
tion due to twisting, and the sidewise deflection due to bending, 
become excessive, and not permissible in good design. This trouble 
is remedied by placing a bearing at some point closer to the disk, 
which, as it decreases L, decreases the bending moment and, there¬ 
fore, the transverse deflection. The angular deflection can be de¬ 
creased only by bringing the resistance and load nearer together. 

Note. The above implies, of course, that the diameter of the shaft is not 
changed, it being obvious that increase of diameter means increase of strength 
and corresponding decrease of both angular and transverse deflection. 

If the speed of the shaft be very high, and the distance be¬ 
tween bearings, represented by L, be very great, the shaft will take 



a shape like a bowstring when it is vibrated, and smooth action 
cannot be maintained. 

It is necessary to carry the cases of Figs. 64 and 65 but a single 
step farther to illustrate the actual working conditions of shafting 
in machines. Suppose the rigid post to have the shaft passing clear 
through it, and to act as a bearing, so that the shaft can freely rotate 

























114 


MACHINE DESIGN 


in it, the resistance being exerted somewhere beyond. The twisting 
moment will be unchanged, also the bending moment; but the effect 
of the bending moment will be on each particle of the shaft in suc¬ 
cession, first putting compression on a given particle, and then 
tension, then compression again, and so on, a complete cycle being 
performed for each revolution. This brings out a very important 
difference between the bending stress in pins and the bending stress 
in rotating shafts. In the one case the bending stress is non-reversing; 
in the other, reversing; and a much higher fiber stress is permissible 
in the former than in the latter. 

Theory. Simple Torsion. In the case of simple torsion, the 
stress induced in the shaft is a shearing one. The external moment 
acts about the axis of the shaft, or is a polar moment; hence in the 
expression for the moment of the internal forces, the polar moment 
of inertia must be used. Now, from mechanics, 



c 


and for circular section of diameter d 

I _ d 3 
c ~ 5.1 


therefore 




from which the diameter for any given twisting moment and fiber 
stress can readily be found. 

For a hollow shaft this expression becomes 

T _ S(d* - Q 
5.1 d 0 



Simple Bending. The stresses induced in a pin or shaft under 
simple bending are compression and tension. The external moment 
in this case is transverse, or about an axis across the shaft; hence 
the direct moment of inertia is applicable to the equation of forces. 



c 


and for circular section of diameter d 







MACHINE DESIGN 


115 


d* 

10.2 


therefore 


M= 


Sd 3 
10.2 


(71) 


For a hollow shaft or pin this expression becomes 

M= ~ ^i 4 ) 

10.2 d n 


(72) 


In the greater number of cases met with in practice, there are 
two or more simple stresses acting at the same time, and, although 
the shaft may be strong enough for any 
one of these alone, it may fail under 
their combined action. The most com¬ 
mon cases are as follows: 

Tension or Pressure Combined with 
Bending. In Fig. 66, the load W pro¬ 
duces a tension acting over the whole 
area of d, due to its direct pull. It also 
produces a bending action due to the 
leverage R, which puts the fibers at B in 
tension and those at the opposite side in 
compression It is evident, therefore, 
that by taking the algebraic sum of the 
stresses at either side the net stress may 
be obtained. It is evident that the 

greatest and also the controlling stress will occur on the side 
where the stresses add, i. e., on the tension side. Hence, from 
mechanics, 

\y~ 



Fig. 66. Tension or Compression 
with Bending 


or due to direct tension 


Also 


S = 


WR = 


4W 

nd 2 


Sd 3 


(73) 


10.2 


or due to bending 
























116 


MACHINE DESIGN 


10.2 WR 

d 3 



Hence, the combined tensional stress acting at the point B, or, in 
fact, at any point on the extreme outside of the vertical shaft toward 
the force W, is 


4TF 10.2 WR 

nd 2 + d 3 



If W acted in the opposite direction, the greatest stress would 
still be at the side B , but instead of a tension, would be a compression 

and of the same magnitude as before. 

Tension or Compression with Torsion. 
In Fig. 67, V might be the end load on a 
vertical shaft; and the two forces W might 
act in conjunction with it, as in the case of 
Fig. 64, at the radius R. This case is not 
very often met with as it is usually possi¬ 
ble to combine the moments, find an 
equivalent moment of a simple kind, and 
use the corresponding simple fiber stress. 
In the case in question a direct stress is 
combined with a shearing stress, and 
Fi g. Tenstaorcompression mec hanics gives the following solution: 

Let S a = simple shearing stress (lbs. per 
sq. in.); S c = simple compressive stress (lbs. per sq. in.); S n = 
resultant shearing stress (lbs. per sq. in.); S tc = resultant compres¬ 
sive stress (lbs. per sq. in.), then 




2IVR = 

S B d 3 



5.1 

or 

8.~ 

5.1(2 WR) 


d 3 

Also 

V = 

7td 2 S e 



4 

or 

S c = 

4 V 



7td 2 

























MACHINE DESIGN 


117 


Now, from a solution given in simplest form in “Merriman’s Me¬ 
chanics”—which the student may consult, if desired—values for 
the resultant stresses may be found. Whichever of these is the crit¬ 
ical one for the material used, should form the basis for its diameter. 

s„ = yl s ‘ 2+ T (78) 

Also 

= T + >i s > + t (79) 


Bending Combined with Torsion. In Fig. 68, the load W 
acts not only to twist the shaft off, but also presses it sidewise against 



the bearings. As it is usually customary to figure the maximum 
moment as taking place at the center of the bearing, the length L, 
which determines the bending moment, is taken to that point. The 
theory of the stress induced in this case is complicated. In order 
to make the magnitude of the moments clearer, let the two equal 
and opposite forces F and F 1 be introduced, each equal to W, at the 
point C. Evidently this can be done without changing the equilibrium 
of the shaft in any way. W and F 1 act as a couple giving a twisting 
moment WR; and F acts with a leverage L, producing a bending 
moment FL = WL , at the middle of the bearing. 

If, now, an equivalent twisting moment, or an equivalent bend¬ 
ing moment is found which would produce the same effect on the 
fibers of the shaft as the two combined, the calculation of the diameter 
can be treated as a simple case, and the procedure would be the 
same as in the cases of simple torsion and simple bending considered 
above. This relation is given in mechanics as 































118 


MACHINE DESIGN 


a . M 
M = T + 


ta( 


m 2 +t 2 


(80) 

(81) 


t % = m+Vm 2 + t 2 

These expressions are true in relation to each other, on the assump¬ 
tion that the allowable fiber stress S is the same for tension, com¬ 
pression, and shearing. For the material of which shafts are usually 
made, this is near enough to the truth to give safe and practical re¬ 
sults. Using the expressions for internal moments of resistance 
as previously noted for circular sections 

Sd 3 


Also 


M 


T = 


10.2 


Sd 3 

5.1 


(82) 


(83) 


Either equation may be used and will give the same value for the 
diameter d. For the sake of simplicity, the torsion equations are 
generally used. 

The expression V M 2 + T 2 is one that would sometimes be a 
tedious task to calculate. By inspection it is readily seen that this 
quantity may be graphically represented by means of a right-angled 
triangle having M and T as the sides. Lay off on a piece of paper to 
some convenient scale, the moments M and T, as the sides of a right- 
angled triangle; then, the measure of the hypothenuse gives the values 

of V M+ T 2 . Even if the drawing is made to a small scale, the 
accuracy of the reading will be sufficient to enable the value for d 
to be solved very closely. 

Deflection. For a shaft subjected to pure torsion, as in Fig. 
64, the angular deflection due to the load may be carried to a certain 
point before the limit of working fiber stress is exceeded. The 
equation worked out from mechanics for this condition, is 

584 TL (84) 


A 0 = 


Gd 4 


which gives the number of degrees of angular deflection for 
a shaft whose modulus of elasticity, torsional moment, and length 
are known. 

Isote. The shearing modulus of elasticity of ordinary shaft steel runs 
from 10,000,000 to 13,000,000, giving as an average about 12,000,000. 









MACHINE DESIGN 


119 


By the well-known relation of “Hooke’s law”—stresses pro¬ 
portional to strains within the elastic limit of the material— 

A° SL 
360° ~ 7 rGd 

or 

o _ A°irGd 

A twist of one degree in a length of twenty diameters is a usual 
allowance. Substituting A = 1, L = 20 d, and G = 12,000,000, 

S = 5,240, nearly (86) 

This is a safe value for shearing fiber stress in steel. In fact, in 
calculations for strength, even for reversing stresses, the usual figure 
is 8,000 (lbs. per square inch), thus indicating that the relation of 
one degree to twenty diameters is well within the limit of strength. 
For a hollow shaft the above formula becomes 



584 TL 
G{d 4 —d 4 ) 



Transverse deflection occurs when the shaft is subjected to a 
bending moment. It may, therefore, exist alone or in conjunction 
with angular deflection. Transverse deflection of shafts, however, 
rarely exists up to the point of limiting fiber stress, because before 
that point is reached the alignment of the shaft is so disturbed that 
it is not practicable as a device for transmitting power. A trans¬ 
verse deflection of .01 inch per foot of length is a common allowance; 
but it is impossible to fix any general limit, as in many cases this figure, 
if exceeded, would do no harm, while in others—such as heavily 
loaded or high-speed bearings—even the figure given might be fatal 
to good operation. 

The formula for transverse deflection, deduced from mechanics, 
varies with the system of loading. The three most common con¬ 
ditions only are given below, reference to the handbook being neces¬ 
sary if other conditions must be satisfied: 

(1) Fixed at one end, loaded at the other, 

WU 








120 


MACHINE DESIGN 


(2) Supported at ends, loaded in middle , 

WU 
e ~ 48 El 

(3) Supported at ends , loaded uniformly , 

_ 5WL 3 

e ~ 384 El 

For transverse deflection the direct modulus of elasticity must 
be used, for the fibers are stretched or compressed, instead of being 
subjected to a shearing action. The most usual value of the direct 
modulus of elasticity for ordinary steel is 30,000,000, and is denoted 
in most books by the symbol E. Both the shearing and direct 
moduli of elasticity are really nothing but the ratio of the stress to 
the strain produced by that stress, it being assumed that the given 
material is perfectly elastic. A material is supposed to be perfectly 
elastic up to a certain limit of stress, and it is within this limit that 
the relation as above holds good. Expressed in the form of an 
equation this would be 

F _ S _SL 

e ~ e (91) 

L 

Centrifugal Whirling. If a line shaft deflects but slightly, due 
to its own weight, or the weight or pressure of other bodies upon it, 
and then be run at a high speed, the centrifugal force set up in¬ 
creases the deflection, and the shaft whirls about the geometrical 
line through the centers of the bearings, causing vibration and wear 
in the adjoining members. It is evident that the practical remedy 
for this tendency in a shaft of given diameter and speed is to locate 
the bearings sufficiently close to render the action of small effect. 

Many formulas might be given for this relation, each being 
based on different assumptions. Perhaps as widely applied and 
as simple as any, is the Rankine formula, which sets the limit of 
length between bearings for shafts not greatly loaded by intermediate 
pulleys or side strains, 

B = 175 J A 

\ N 

Horse-Power of Shafting. Horse-power is a certain specific 



(89) 

(90) 






MACHINE DESIGN 


121 


rate of doing work, viz, 33,000 foot-pounds per minute. Hence, 
to find the horse-power that a shaft will transmit, first, find the work 
done, and then relate it to the speed. Take, for example, the case 
of a pulley, using the symbols P = driving force at rim of pulley 
(lbs.); R = radius of pulley (inches); N = number of revolutions 
per minute; and H = horse-power. Then 


or 


Work = force X distance = P X (2 nRN) 


2tcPRN 

33000X12 


and PR = 


63025 H 
N 



This is one of the most useful equations for calculations involving 
horse-power. By it the number of inch-pounds torsion for any 
horse-power can be at once ascertained. 

It should be clearly noted, however, that in this equation the 
bending moment does not enter at all. Hence any shaft based in 
size on horse-power alone, is based on torsional moment alone, bend¬ 
ing moment being entirely neglected. In many cases the bending 
moment is the controlling one as to limiting fiber stress. Hence 
empirical shafting formulas depending upon the horse-power rela¬ 
tion are unsafe, unless it is definitely known just what torsional 
and bending moments have been assumed. 

The only safe way to figure the size of a shaft is to find accurately 
what torsional moment and bending moment it has to sustain, and 
then combine them for the equivalent moment, introducing the 
element of speed as basis for assumption of a high or low working 
fiber stress. 

Practical Modification. The practical methods of handling 
the theoretical shaft equations have reference to the fit of the shaft 
within the several pieces upon it. The running fit of a shaft in a 
bearing is usually considered to be so loose that the shaft could freely 
deflect to the center of the bearing. This is doubtless an extreme 
view of the case, but it is the only safe assumption. Hence a shaft 
running in bearings, Fig. 69, is supposed to be supported at the 
centers of those bearings, and its theoretical strength is based on 
this supposition. 

For a tight or driving fit upon the shaft, a safe assumption to 
make is that there is looseness enough at the ends of the fit to per¬ 
mit the shaft to be stressed by the load a short distance within the 




122 


MACHINE DESIGN 


faces of the hub, say from \ inch to 1 inch. Referring to Fig. 69, 
suppose Pj to be the transverse load, exerted through a hub fast 
upon the part of the shaft d 3 . Taking moments about the center of 
one bearing, and solving for the reaction at the center of the other, 

P x u = R x K 


or 










R> 

Pj u 

K 


(94) 

Also 









pj 

= R,K 



or 










r 2 

_P,t 

IC 


(95) 

Now, as 

far as 

the part 

of shaft 

d 3 is concerned, it 

may 

depend for 

its size 

on the 

bending 

moment 

R 2 b, or on R^a. 

The 

reason the 



lever arm is not taken to the point directly under the load P v is 
because it is not practically possible to break the shaft at that point 
on account of the reinforcement of the hub, which is tightly fitted 
upon it. Trying these moments to see which is the greater, it is 
found that the greater moment always occurs in connection with 
the longer lever arm. Hence R 2 b will be greater than R x a. Writing 
the equation of external moment = internal moment 

Sd* 













































MACHINE DESIGN 


123 


or 


For the size of bearing A , the maximum bending moment is 

r A = sd < 3 
1 2 10.2 




10.2 

2S 



or 


For the size of bearing B, the maximum moment is 

ft 4 = Sd* 

2 2 10.2 




10.2 R 2 L 2 
2S 



Note. The above calculations are, of course, on the assumption that 
no torsion is transmitted either way through this axle. In that case com¬ 
bined torsion and bending occur. This has been made sufficiently clear in 
preceding paragraphs and in Part I, to require no further illustration. 

The dotted line in Fig. 69 shows the theoretical shape the axle 
should take under the assumed conditions. The practical modifi¬ 
cation of this shape is obvious. At the shoulders of the shaft the 
corners should not be sharp, but carefully filleted, to avoid the pos¬ 
sible starting of a crack at those points. 

Often the diameter of certain parts of a shaft may be larger 
than strength actually calls for. For example, in Fig. 69, the part 
d 3 need only be as large as the dotted line; but it is obvious that unless 
the key is sunk in the body of the shaft, the hub could not be slipped 
into place over the part d A . If, however, the diameter d 3 be made 
large enough so that the bottom of the key will clear d v the rotary 
cutter which forms the key way in d 3 will also clear d v and the key 
way can be more easily produced. 

In cases where fits are not required to be snug, a straight shaft 
of cold-rolled steel is commonly used. Here any parts fastened on 
the middle of the shaft have to be driven over a considerable length 
of the shaft before they reach their final position. Moreover, there 
is no definite shoulder to stop against, and measurement has to be 
resorted to in locating them. 

It does not pay to turn any portion of a cold-rolled shaft, unless 
it be the very ends, as the release of the “skin tension” in such material 










124 


MACHINE DESIGN 


is sure to throw the shaft out of line and necessitate subsequent 
straightening. 

Turned-steel shafts for machines may with advantage be slightly 
varied in diameter wherever the fit changes; and although the pro¬ 
duction of shoulders costs something, yet it assists greatly in bring¬ 
ing the parts to their exact location, and enables the workman to 
concentrate his best skill on the fine bearing fits, and to save time 
by rough-turning the parts that have no fits. 

Hollow shafts are practicable only for large sizes. The ad¬ 
vantages of removing the inner core of metal, aside from some specific 
requirement of the machine, are that it eliminates all possibility of 
cracks starting from the checks that may exist at the center, permits 
inspection of the material of a shaft, and, in case of hollow-forged 
shafts, gives an opening for the forging mandrel. In the last case, 
the material is improved by a rolling process. 

The material most common for use in machine shafting is the 
ordinary machinery steel, made by the Bessemer process. This 

steel is apt to be “seamy,” 
and often contains checks 
and flaws that are de¬ 
tected only upon sudden 
and unexpected breakage 
of a part apparently 
sound. This character¬ 
istic is a result of the 
process employed in the 
manufacture of the steel, 
and thus far has never 
been wholly eliminated. 
Bessemer steel is, never¬ 
theless, a very useful material, and the above weakness is not so 
serious but that this kind of steel may be used with success in the 
great majority of cases. 

When a more homogeneous shaft is desired, open-hearth steel 
is available. This is a more reliable material to use than the Bes¬ 
semer, and costs somewhat more. It makes a stiff, true, fine-sur¬ 
faced shaft, high-grade in every respect. It is usually specified for 
armature shafts of dynamos and motors. 































MACHINE DESIGN 


125 


Steels of special strength, toughness, and elasticity are made 
under numerous processes, of which nickel steel is perhaps the most 
conspicuous example. While this steel is very expensive, yet its 
great strength, in connection with 
other excellent qualities, makes 
it an extremely valuable material, 
wherever light weight is essential, 
or contracted space demands small 
size. 

Example. The overhung 
crank of a steam engine shown 
in Fig. 70 has a force of 32,000 
lbs. acting at the center of the crank pin. Assume S equal to 10,000 
lbs. per sq. in. and calculate the diameter of the crank shaft. 

Solution. This is an example of combined bending and torsion. 
There is torsion on the crank shaft due to a force 32,000 lbs. acting 
at the end of an arm 10 inches in length. The bending is also due to 
a force 32,000 lbs. acting on an arm 6 inches in length. 

M + VM 2 + T 2 
M 



Fig. 71. Triangle of Forces 


T. 


M 

T 


2 +-Lvm 2 + T 2 

6 X 32000 = 192000 
10 X 32000 = 320000 


The value VM 2 + T 2 may be found graphically by using a right 
triangle. Then drawing a right triangle to a scale such that one leg 
is proportional in length to 192,000 and the other to 320,000, as in 
Fig. 71, measure the hypothenuse and this will be found equal to 
373,000 about. 

T e = 192000 + 373000 = 565000 
192000 . 373000 


M = 


+ 


= 282500 


2 2 

By substituting the values for M 9 and T„ in equations 82 and 
83, respectively, the same result for the diameter is obtained. 

Sd 3 


M = 


10.2 











126 


MACHINE DESIGN 


282500 = 


lOOOOd 3 

10.2 


282500 X 10.2 

10000 




d 3 = 289.56 
d =6.6, say 6f" 

PROBLEMS FOR PRACTICE 

1. Required the twisting moment on a shaft that transmits 
30 horse-power at 120 revolutions per minute. 

2. Find the diameter of a steel shaft designed to transmit 50 
horse-power at 150 revolutions per minute. 

3. Assuming same data as in Problem 1, find the diameters 
of a hollow shaft for a value of S = 8,000. 

4. A belt on an idler pulley embraces an angle of 120 degrees. 
Assuming tension of belt 1,000 pounds on each side, and pulley 
located midway between bearings—which are 30 inches from center 
to center—what is the diameter of shaft required? 

5. Calculate the diameter of a steel shaft designed to transmit 
a twisting moment of 400,000 inch-pounds and also to take a bending 
moment of 300,000 inch-pounds. 

6. Find the angular deflection in a 4-inch shaft 20 feet long 
when subjected to a load of 5,500 pounds applied to an arm of 30-inch 
radius. Assume transverse modulus of elasticity equal to 12,000,000. 


SPUR GEARS 

NOTATION—The following notation is used throughout the chapter on 

Spur Gears: 


b = Breadth of rectangular section 
of arm (inches) 

C = Width of arm extended to pitch 
line (inches) 

c = Distance from neutral axis to 
outer fiber (inches) 

D = Pitch diameter of gear (inches) 
F = Face of gear (inches) 

/ = Clearance of tooth at bottom 

(inches) 

G = Thickness of arm extended to 
pitch line (inches) 

H = Thickness of tooth at any sec¬ 
tion (inches) 


/ 1 = Coefficient of friction between 
teeth 

N = Number of teeth 
n = Number of arms 
P = Diametral pitch (teeth per inch 
of diameter) 

P 1 = Circular pitch (inches) 

Q, = Normal pressure between 

teeth (lbs.) 

R, R x = Resultant pressure between 

teeth (lbs.) 

r, r, = Radius of pitch circles (inches) 
S = Fiber stress of material (lbs. per 
sq. in.) 




MACHINE DESIGN 


127 


h = Depth of rectangular section of s = Addendum of tooth (inches) = 
arm (inches) Dedendum of tooth 

I = Moment of inertia t = Thickness of tooth at pitch line 

K = Thickness of rim (inches) (inches) 

L = Distance from top of tooth to W = Load at pitch line (lbs.) 

any section (inches) y =. Coefficient for “Lewis” formula 

M, M, = Revolutions per minute 

Analysis. If a cylinder be placed on a plane surface, with 
its axis parallel to the plane, an attempt to rotate the cylinder about 
its axis would cause it to roll on the plane. 

Again, if two cylinders be provided with axial bearings, and 
be slightly pressed together, motion of one about its axis will cause 
a similar motion of the other, the two surfaces rolling one on the other 
at their common tangent line. If moved with care, there will be no 
slipping in either of the above cases—which is explained by the fact 
that no matter how smooth the surfaces may appear to be, there is 
still sufficient roughness to make the little irregularities interlock 
and act like minute teeth. 

The magnitude of the force possible to be transmitted depends 
not only on the roughness of the surfaces, but on the amount of 
pressure between them. Suppose that one cylinder is a part of a 
hoisting drum, on which is wound a rope with a weight attached. 
The weight can be made so great that, no matter how hard the 
two cylinders are pressed together, the driving cylinder will not turn 
the hoisting cylinder, but will slip past it. If now, instead of increas¬ 
ing the pressure, which is detrimental both to cylinders and bearings, 
the coarseness of the surfaces is increased, or, in other words, teeth of 
appreciable size are put on these surfaces, the desired result of posi¬ 
tively driving without excessive side pressure is attained. 

These artificial projections, or teeth, must fit into one another; 
hence, the surfaces of the original cylinders, which have been broken 
up into the alternate projections and hollows, have technically disap¬ 
peared but nevertheless exist as ideal or imaginary surfaces. These 
roll together with the same surface velocities as if in bodily form, 
provided that the curves of the teeth are correctly formed. Several 
mathematical curves are available for use as tooth outlines, but in 
practice the involute and cycloidal curves are the only ones used for 
this purpose. 

The ideal surfaces are known as 'pitch cylinders or pitch circles. 


128 


MACHINE DESIGN 


In Fig. 72 is shown an end view of such a pair of cylinders in contact 
at their pitch point P. In gear calculations it is assumed that there 
is no slip between the pitch circles acting as driving cylinders; hence 
the speeds of the two pitch circles at the pitch point are equal. If 
M and M 1 be the revolutions per minute of the cylinders, respectively, 
r and r 1 their radii, then 

2tt r M = 2irr x M t 

or 

M r t 
M~ r 

That is, the number of revolutions varies inverselv as the radii. 

The simple calculation as above is the key to all calculations 
involving gear trains in reference to their speed ratio. 




Fig. 72. Pitch Cylinders and Teeth 



The distance from the center of one tooth to the center of the 
next measured along the pitch circle is called the circular pitch. The 
ratio of the number of teeth per inch of diameter is the diametral 
pitch. The relation of the diametral pitch, circular pitch, number of 
teeth, and pitch diameter is expressed in the following equations 

3.1416 

P 1 

d = A (ioi) 

Fig. 73 represents cycloidal teeth in the two extreme positions 















MACHINE DESIGN 


129 


of beginning and ending contact. The normal pressure Q or 
between the teeth in each position acts through the pitch point 0, 
as it must always do in order to insure the condition of ideal rolling 

of the pitch circles and the velocity ratio proportional to —. As 

r 

the surfaces of the teeth slide together, frictional resistance is pro¬ 
duced at their point of contact. This force is widely variable, depend¬ 
ing on the material and condition 
of the tooth surfaces, whether 
smooth and well lubricated, or 
rough and gritty. As this resistance 
acts in conjunction with the normal 
force between the teeth, a paral¬ 
lelogram of forces may be con¬ 
structed on these two as a base, 
the resultant pressure between the 
teeth being slightly changed there¬ 
by, as shown in Fig. 73. It is true, 
however, that in nearly all cases in Flg ' 73 ‘ °y cl0ldaI Teeth 

practice, the bending stress is the controlling one from a theoretical 
standpoint. Morover, the designer must consider the form and 
strength of the tooth when it is under the condition of maximum 
moment. This evidently occurs at the beginning of contact, for the 
follower teeth; and at the end of contact, for the driver teeth. In 
the particular case illustrated in Fig. 73, if the material in both 
gears were the same, tooth C, being the weaker at the root, would 
probably break before B; but if C were of steel, and B of cast iron, 
B might break first. 

It will be noticed that R is nearly parallel to the top of the tooth; 
and it may easily happen that the friction may become of such a 
value that it will turn the direction of R until it lies along the top of 
the tooth exactly, which is the condition for maximum moment. 
For strength calculations it is usual to consider this condition as 
existing in all cases. 

At the beginning of contact there is more or less shock when 
the teeth strike together, and this effect is much more evident at 
high speeds. There is also at the beginning of contact a sort of 
chattering action as the driving tooth rubs along the driven tooth. 










130 


MACHINE DESIGN 


Uniform distribution of pressure along the face of the tooth is 
often impaired by uneven wear of the bearings supporting the gear 
shafts, the pressure being localized on one corner of the tooth. The 
same effect is caused by the accidental presence of foreign material 
between the teeth. Again, in cast gearing, the spacing may be 
irregular, or, on account of draft on the pattern, the teeth may bear 
at the high points only. While it is usual to consider that the load 
is evenly distributed along the face of the tooth, yet the above con¬ 
siderations show that an ample margin of strength must always be 
allowed on account of these uncertainties. 

When the number of teeth in the mating gears is high, the load 
will be distributed between several teeth; but, as it is almost certain 

that at some time the proper distribution 
of load will not exist, and that one tooth 
will receive the full load, it is considered 
that practically the only safe method is so 
to design the teeth that a single tooth may 
be relied upon to withstand the full load 
without failure. 

Theory. Based on the analysis as given, 
the theory of gear teeth assumes that one 
tooth takes the whole load, and that this 
load is evenly distributed along the top of 
the tooth and acts parallel with its base, thus reducing the condition 
of the tooth to that of a cantilever beam. The magnitude of this 
load at the top of the tooth is assumed to be the same as the force 
transmitted at the pitch circle. This condition is shown in Fig. 74. 
Equating the external moment to the internal moment 



Fig. 74. Tooth as a Beam 


SI SFH 3 

WL ~ c ~ 6 


( 102 ) 


The thickness II is usually taken either at the pitch line or at the 
root of the tooth just before the fillet begins; and L, of course, is 
dependent on the tooth dimensions. The formula is most readily 
used when the outline of the tooth is either assumed or known, a 
trial calculation being made to see if it will stand the load, and a 
series of subsequent calculations followed out in the same way until 
a suitable tooth is found. This method is pursued because there 







MACHINE DESIGN 


131 


are certain even pitches which it is desirable to use; and it is safe to 
say that any calculation figured the reverse way would result in 
fractional pitches. The latter course may be used, however, and 
the nearest even pitch chosen as the proper one. 

As stated in the analysis, there are a great many circumstances 
attending the operation of gears which make impossible the purely 
theoretical application of the beam formulas. For this reason there 
is no one element of machinery which depends so much on experience 
and judgment for correct proportion as the tooth of a gear. Hence 
it is true that a rational formula based on the theoretical one is really 
of the greater practical value in tooth design. 

If formula 102 is examined it is found that when solved for W 


W = 


SFH 2 
6 L 


(103) 


Of these quantities, II and L are the only variables, the others being 
given any desired value. H and L depend upon the circular pitch 



P 1 and the curvature and outline of the tooth. If now a standard 
system of teeth could be settled upon, a coefficient could be estab¬ 
lished to be used to take the place of the variable part of H and L, 
which depends on the outline of tooth, and thus an empirical formula 
which would be on a theoretical basis would be obtained. This, 
Wilfred Lewis has done; and this formula is more universally 
used and with more satisfactory practical results than any other 
formula, theoretical or practical, that has ever been devised. His 
coefficient is known as y, and was determined from many actual 
drawings of different forms of teeth showing the weakest section. 
This coefficient is worked out for the three most common systems as 
follows: 


y 


0.154 


0.912 


For 20° involute, 


N 


(104) 










132 


MACHINE DESIGN 


For 15° involute, 

y = 0.124 - 

0.684 

(105) 

and cycloidal, 

N 

For radial flanks, 

y = 0.075 - 

0.276 

(106) 

N 

Note. The tooth 

upon which the above is based is the American stand 


ard, or Brown & Sharpe tooth, for which the proportions are shown in Fig. 75. 
The “Lewis” formula* is 


W = SP 1 Fy (107) 

In Table VIII may be found the value of S for different speeds. 

TABLE VIII 

* 

Safe Working Stresses for Different Speeds 



The usual method of handling these formulas is as follows: 

The pitch circles of the proposed gears are known or may be assumed; 
hence W is readily figured, as well as the speed of the teeth, whence S may be 
found in Table VIII. The desired relation of F to P 1 may be arbitrarily 
chosen, when P 1 and y become the only unknown quantities in the equation. 
A shrewd guess may be made for the number of teeth, and y calculated there¬ 
from. Then solve the equation for P 1 , which will undoubtedly be fractional. 
Choose the nearest even pitch, or, if it is desired to keep an even diametral 
pitch, the fractional pitch that will bring an even diametral pitch. Now, 
from this final and corrected pitch, and the diameter of the pitch circle, cal¬ 
culate the number of teeth N in the gear. Check the assumed value of y by 
this positive value of N. 

Another good way of using this formula is to start with the pitch, 
and the face desired, and the diameter of the pitch circle. In this 

*A full and convenient statement of the Lewis formula will be found in “Kent's 
Mechanical Engineers’ Pocket-Book.” 






































MACHINE DESIGN 


133 


case W is the only unknown quantity, and when found should be 
compared with the load required to be carried. If it is too small, 
make another and successive calculations until the result approximates 
the required load. 

RIM, ARMS, AND HUB 

Analysis. The rim of a gear has to transmit the load on the 
teeth to the arms. It is thus in tension on one side of the teeth in 
action, and in compression on the other. The section of the rim, 



however, is so dependent on other practical considerations which 
call for an excess of strength in this respect, that it is not considered 
worth while to attempt a calculation on this basis. 

Gears seldom run fast enough to make necessary a calculation 
for centrifugal force; and, in general, it may be said that the design 
of the rim is entirely dependent on practical considerations. 

Theory. The arms of a gear are stressed the same as pulley 
arms, the same theory answering for both, except that a gear rim 
always being much heavier than a pulley rim, the distribution of load 
amongst the arms is better in the case of a gear than of a pulley, and 
it is usually safe to assume that each arm of a gear takes its full pro¬ 
portion of load; or, for an oval section, equating the external moment 
to the internal moment as in the case of pulleys, 

WD 

—-—= 0.0393 Sh* (110) 

Heavy spur gears have the arms of a cross- or T-section, Fig. 
76, the latter being especially applicable to the case of bevel gears 
where there is considerable side thrust. The simplest way of treat- 




















134 


MACHINE DESIGN 


ing such sections is to consider that the whole bending moment is 
taken by the rectangular section whose greater dimension is in the 
direction of the load. The rest of the section, being close to the 
neutral axis of the section, is of little value in resisting the direct 
load, its function being to give sidewise stiffness. The equation for 
the cross- or T-style of arm, then is 


1L D SW 

n ^ 2 6 


( 111 ) 


Either b or h may be assumed, and the other determined. As a 
guide to the section, b may be taken at about the thickness of the 
tooth. 

Gear hubs are in no wise different from the hubs of pulleys or 
other rotating pieces. The depth necessary for providing sufficient 



Fig. 77. 



Mortise Teeth 


strength over the key to avoid splitting is the guiding element, and 
can usually be best determined by careful judgment. 

Practical Modification. The practical requirements, which 
no theory will Satisfy, are many and varied. Sudden and severe 
shock, excessive wear due to an atmosphere of grit and corrosive 
elements, abrupt reversal of the mechanism, the throwing-in of 
clutches and pawls, the action of brakes—these, and many other 
things have an important influence on gear design, but not one of 
them can be calculated. The only method of procedure in such 
cases is to base the design on analysis and theory as previously given, 
and then add to the face of gear, thickness of tooth, or pitch an amount 
which judgment and experience dictate as sufficient. 
































MACHINE DESIGN 


135 



Excessive noise and vibration are difficult to prevent at high 
speeds. At 1,000 feet per minute, gears are apt to run with an un¬ 
pleasant amount of noise. At speeds beyond this, it is often neces¬ 
sary to provide mortise teeth, or teeth of hard wood set into a cast- 
iron rim, Fig. 77. Rawhide pinions are useful in this regard. Fine 


Fig. 78. Full-Shrouding Fig. 79. Half-Shrouding 

pitches with a long face of tooth run much more smoothly at high 
speeds than a coarse pitch and narrow-faced tooth of equal strength. 
Greater care in alignment of shafts, however, is necessary, also stiffer 
supports. 

Should it be impracticable to use a standard tooth of sufficient 
strength there are several ways in which the carrying capacity can be 
increased without increasing the pitch, viz, use a stronger material, 
such as steel; shroud the teeth; use a hook tooth; use a stub tooth. 

Shrouding a tooth consists in connecting the ends of the teeth 
with a rim of metal. When this rim is extended to the top of the 
tooth, the process is called full-shrouding, Fig. 78, and when carried 
only to the pitch line, it is termed half-shrouding, Fig. 79. The 




































































136 


MACHINE DESIGN 


theoretical effect of shrouding is to make the tooth act like a short 
beam built in at the sides; and the tooth will practically have to be 
sheared out in order to fail. This modification of gear design re¬ 
quires the teeth to be cast, as the cutter cannot pass through the 
shrouding. The strength of the shrouded gear is estimated to be 
from 25 to 50 per cent above that of the plain-tooth type. 

The hook-tooth gear, Fig. 80, is applicable only to cases where 
the load on the tooth does not reverse. The working side of the 
tooth is made of the usual standard curve, while the back is made 
of a curve of greater obliquity, resulting in a considerable increase 
of thickness at the root of the tooth. A comparison of strength 

LIGHT PRESSURE 
ON BACK OF TOOTH. 



between this form and the standard may be made by drawing the 
two teeth for a given pitch, measuring their thickness just at top of 
the fillet, and finding the relation of the squares of these dimensions. 
The truth of this relation is readily seen from an inspection of the 
formula of bending moment, equation (88). 

The stub tooth merely involves the shortening of the height 
of the tooth in order to reduce the lever arm on which the load acts, 
thus reducing the moment, and thereby permitting a greater load 
to be carried for the same stress. 

The rim of a gear is dependent for its proportions chiefly on 
questions of practical moulding and machining. It must bear a 
certain relation to the teeth and arms, so that, when it is cooling in 
the mould, serious shrinkage stresses will not be set up, forming 
pockets and cracks. Moreover, when under pressure of the cutter 
in the producing of the teeth, it must not chatter or spring. This 
condition is quite well attained in ordinary gears when the thick¬ 
ness of the rim below the base of the tooth is made about the same 
as the thickness of the tooth. 

The stiffening ribs and arms must all be joined to the rim by 



MACHINE DESIGN 


137 


TABLE IX 
*Gear Design Data 


Diametral. 

P 

n 

•j 

If 

2 

to 

3 

31 

4 

5 

6 

8 

Face. 

F 

61 

51 

41 

3f 

31 

2f 

21 

21 

If 

11 

Thickness of arm 
when extended 
to pitch line. 

G 

If 

H 

h 

1 

1 

% 

a 

4 

% 

5. 

8 

1 

Width of arm when 
extended to pitch 
line. 

C 

4 

3f 

3 

21 

21 

2 

If 

11 

If 

H 

Thickness of rim .. . 

K 

2| 

2f 

21 

If 

H 

If 

H 

1 

f 

f 

Depth of rib. 

E 

2 

If 

11 

H 

1 

I 

f 


1 

f 

Thickness of web. . . 

T 

If 

1 

1 

f 

5. 

8 

% 

1 

% 

f 

% 


* Measurements given in inches. Letters refer to Fig. 81 
Number of arms, 6. 

Give inside of rims and hub a draft of £ inch per foot. 


ample fillets, and the cross-section must be as uniform as possible, 
to prevent unequal cooling and consequent pulling-away of the 



arms from the rim or hub. Often the calculated size of the arms 
at both rim and hub has to be modified considerably to meet this 
requirement. 


























































































































138 


MACHINE DESIGN 


The arms are usually tapered to suit the designer’s eye, a small 
gear requiring more taper per foot than a large one. Both rim and 
hub should be tapered J inch per foot to permit easy drawing-out 
from the mould. 

The proportions given in Table IX have been used with success 
as a basis of gear design in manufacturing practice. The table 
will serve as an excellent guide in laying out, and can be closely 
followed, in most cases with but slight modification. Web gears are 
introduced for small diameters where the arms begin to look awkward 
and clumsy. 

BEVEL GEARS 

Analysis. It is possible to consider bevel gears as the general 
case of which spur gears are a special form. The pitch surfaces of 
spur gears described as cylinders when mathematically considered, 
are cones whose vertices are infinitely distant, while bevel gears like¬ 
wise are based on pitch cones, but with a vertex at some finite point, 
common to the mating pair. Hence, as might be expected, the laws 
of tooth action are similar in bevel gears to those in the case of 
spur gears. The profile of the tooth in the former case, however, 
is based, not on the real radius of the pitch cone, but on the radius 
of the normal cone; and in the development of the outline, the latter 
is treated just as though it were the radius of a spur gear. The tooth 
thus formed is wrapped back upon the normal cone face, and becomes 
the large end of the tapering bevel-gear tooth. 

The teeth of bevel gears, being simply projections with bases on 
the pitch cones, have a varying cross-section decreasing toward the 
vertex; they also have a trapezoidal section of root, the latter section 
acting as a beam section to resist the cantilever moment due to the 
tooth load. 

The arms must, as in the case of spur gears, transmit the load 
from the tooth to the shaft; in addition, the arms of a bevel gear 
are subjected to a side thrust due to the wedging action of the cones. 
Hence, sidewise stiffness of the arms is more essential in this type of 
gear than in the case of the spur gear. 

Theory. It is evident that the calculations of tooth strength 
based on a trapezoidal section of root would be somewhat compli¬ 
cated; also that the trapezoid in most cases would be but little dif- 


MACHINE DESIGN 


139 


ferent from a true rectangle. Hence, the error will be but slight 
if the average cross-section of the tooth be taken to represent its 
strength, and the calculation made accordingly. 

Fig. 82 shows a bevel-gear tooth with the average cross-section 
in dotted lines. For the purpose of calculation, the assumption is 
made that the section A is called the full length 
of the face of the gear, and that the load which 
this average tooth must carry is the calculated load 
at the pitch line of section A. This is equivalent 
to saying that the strength of a bevel-gear tooth is 
equal to that of a spur-gear tooth which has the 
same face, and a section identical with that cut 
out by a plane at the middle of the bevel tooth. 

The load, as in the case of the spur gear, should Fig. 82 ^, 0 ^® vel Gear 
be taken at the top of the tooth; and its magnitude 
can be conveniently calculated at the mean pitch radius of the bevel 
face, without appreciable error. 

This similarity to spur gears being borne in mind, the 
calculation for strength needs no further treatment. Once the 
average tooth is assumed or found by layout, a strict following- 
out of the methods pursued for spur-gear teeth will bring consis¬ 
tent results. 

Practical Modification. The practical requirements to be met 
in transmission of power by bevel gears are the same as for spur gears; 
but in the case of bevel gears even greater care is necessary to provide 
stiffness, strength, true alignment, and rigid supports. As far as 
the gears themselves are concerned, a long face is desirable; but it 
is much more difficult to gain the advantage of its strength than in 
the case of spur gears, because full bearing along the length of the 
tooth is hard to guarantee. 

The rim usually requires a series of ribs running to the hub 
to give required stiffness and strength against the side thrust which 
is always present in a pair of bevel gears. Instead of arms, the ten¬ 
dency of bevel-gear design, except for very large gears, is toward a 
web on account of the better and more uniform connection thereby 
secured between rim and hub. This web may be lightened by a 
number of holes, so that the resultant effect is that of a number o" 
wide and flat arms. 




140 


MACHINE DESIGN 


The hubs naturally have to be fully as long as those of spur 
gears, because there is greater tendency to rock on the shaft, due 
to the side thrust from the teeth. 

The teeth on small gears are cut with rotary cutters, at least 
two finishing cuts being necessary, one for each side of the taper¬ 
ing tooth. A more accurate method is to plane the teeth on a special 
gear planer, a method which is followed on all gears of any con¬ 
siderable size. The practical requirement here is that no portion 
of the hub shall project so as to interfere with the stroke of the planer 
tool. The requirements of gear planers vary somewhat in this regard. 

Finally, after all that is possible has been done in the design 
of the gear itself to render it suitable to withstand the varied stresses, 
especial attention must be paid to the rigidity of the supporting 
shafts and bearings. Bearings should always be close up to the 
hubs of the gears, and, if possible, the bearing for both pinion and 
gear should be cast in the same piece. If this is not done, the ten¬ 
dency of the separate bearings to get out of line and destroy the full 
bearing of the teeth is difficult to control. Thrust washers are desir¬ 
able against the hubs of both pinion and gear; also proper means of 
well lubricating the same. 

With these considerations carefully met, bevel gears are not the 
bugbear of machine design that they are sometimes claimed to be. 
The common reason why bevel gears cut and fail to work smoothly, 
is that the gears and supports are not designed carefully enough in 
relation to each other. This is also true of spur gears, but the bevel 
gear will reveal imperfections in its design far the more quickly of 
the two. 


WORM AND WORM GEAR 

NOTATION—The following notation is used throughout the chapter on 

Worm and Worm Gear: 

D = Pitch diameter of gear (inches) P l = Circular pitch = Pitch of worm 
E = Efficiency between worm shaft thread (inches) 

and gear shaft (per cent) R = Radius of pitch circle of worm 

/ = Clearance of tooth at bottom (in.) gear (inches) 

i = Index of worm thread (1 for s = Addendum of tooth (inches) 

single, 2 for double, etc.) T = Twisting moment on gear shaft 
L = Lead of worm thread (inches) (inch-lbs.) 

M = Revolutions of gear shaft per T w = Twisting moment on worm shaft 
minute (inch-lbs.) 


MACHINE DESIGN 


141 


M w = Revolutions of worm shaft per t = Thickness of tooth at pitch line 
minute (inches) 

N = Number of teeth in gear W = Load at pitch line (lbs.) 

Analysis. The simplest way of analyzing the case of the worm 
and worm gear is to base it upon an ordinary screw and nut. Take 
for example, the lead screw of a common lathe. The carriage carries 
a nut through which the lead screw passes. By the rotation of the 
screw, the carriage, being constrained by the guides to travel length¬ 
wise of the ways, is moved. This motion is, for a single-threaded 
screw, a distance per revolution equal to the lead of the screw. 

Now, suppose that the carriage, instead of sliding along the 
ways, is compelled to turn about an axis at some point below the 
ways. Also, suppose the top of the nut to be cut off, and its length 
made endless by wrapping it around a circle struck from the center 
about which the carriage rotates. This reduces the nut to a peculiar 
kind of spur gear, the partial threads of the nut now having the 
appearance of twisted teeth. 

This special form of spur gear, based on the idea of a threaded 
nut, is known as a worm gear, and the screw is termed a worm . 
The teeth are loaded similarly to those of a spur gear, but with the 
additional feature of a large amount of sliding along the tooth sur¬ 
faces. This, of course, means considerable friction; and it is, in 
fact, possible to utilize the worm and worm gear as an efficient de¬ 
vice only by running the teeth constantly in a bath of oil. Even 
then the pressures have to be kept well down to insure the required 
term of life of the tooth surfaces. 

It is evident that for one revolution of a single-threaded worm, 
one tooth of the gear will be passed. The speed ratio between the 
worm gear and worm shaft will then be equal to the number of 
teeth in the gear, which is relatively great. Hence the worm and 
worm gear are principally useful in giving large speed reduction in 
a small amount of space. 

Theory. The theory of worm-wheel teeth is complicated 
and obscure. The production of the teeth is simple, a dummy worm 
with cutting edges, called a hob, being allowed to carve its way into 
the worm-gear blank, thus producing the teeth and at the same time 
driving the worm gear about its axis. 

It is clear that if the torsional moment on the worm-gear shaft 


142 


MACHINE DESIGN 


and the pitch radius of the worm gear are known, the load on the 
teeth at the pitch line can be found by dividing the former by the 
latter. Expressed as an equation 

WR = T 
or 

W = L ( 112 ) 

How this value of W is distributed on the teeth, is a question 
difficult to answer. The teeth not only are curved to embrace the 
worm, but are twisted across the face of the gear, so that it would 
be practically impossible to devise a purely theoretical method of 
exact calculation. The most reasonable thing to do is to assume 
the teeth as being equally as strong as spur-gear teeth of the same 
circular pitch, and to figure them accordingly. It is probably true, 
however, that the load is carried by more than one tooth, especially 
in a hobbed wheel; so it will be safe to assume that two—and, in 
case of large wheels, three—teeth divide the load between them. 
With these considerations borne in mind, the case reduces itself to 
that of a simple spur-gear tooth calculation, which has already been 
explained under “Spur Gears.” 

The worm teeth are probably always stronger than the worm- 
gear teeth; so no calculation for their strength need be made. 

The twisting moment on the worm shaft is not determined so 
directly as in the case of spur gears. The relative number of revolu¬ 
tions of the two shafts depends upon the lead of the worm thread 
and the number of teeth in the gear. Lead L is the distance parallel 
to the axis of the worm which any point in the thread advances in 
one revolution of the worm. Pitch P 1 is the distance parallel to the 
axis of the worm between corresponding points on adjacent threads. 
The distinction between lead and pitch should be carefully observed, 
as the two are often confounded, one with the other. 

The thread may be single, double, triple, etc., the index of the 
thread i, being 1, 2, 3, etc., in accordance therewith. The relation 
between lead and pitch may then be expressed by the equation 

L = iP l (113) 

When the index of the thread is changed the speed ratio is 
changed, the relation being shown by the equation 


MACHINE DESIGN 


143 


M 

M W ~N 


(114) 


If the efficiency were 100 per cent between the two shafts, the 
twisting moments would be inversely as the ratio of the speeds, thus 


or 


Ll = JL = ± 

T M w N 



(115) 


but for an efficiency E the equation would be 


i 

T EN 

or 



(116) 


The diameter of the worm is arbitrary. A change of this diam¬ 
eter has no effect on the speed ratio but has a slight effect on the 
efficiency, the smaller worm giving a little higher efficiency. The 
diameter of the worm runs 
ordinarily from 3 to 10 times 
the circular pitch, an average 
value being 4P 1 or 5P 1 . 

A longitudinal cross-sec¬ 
tion through the axis of the 

-T lg. oo. xvaCK 

worm cuts out a rack tooth, and 

this tooth section is usually made of the standard 15° involute form, 
shown in Fig. 83, for a rack. 

The end thrust, of a magnitude practically equal to the pressure 
between the teeth, has to be taken by the hub of the worm against the 
face of the shaft bearing. A serious loss of efficiency from friction 
is likely to occur here. This is often reduced, however, by roller or 
ball bearings. With two worms on the same shaft, each driving into 
a separate worm gear, it is possible to make one of the worms right- 
hand thread, and the other left-hand, in which case the thrust is self- 
contained in the shaft itself, and there is absolutely no end thrust 
against the face of the bearing. This involves a double outfit through¬ 
out, and is not always practicable. 

















144 


MACHINE DESIGN 


There are few mathematical equations necessary for the dimen¬ 
sioning of a worm and worm gear. The formulas for the tooth parts 
as given on pages 131 and 132 apply equally well in this case. 

Practical Modification. The discussion of the efficiency E of 
the worm and worm gear is more of a practical than of a theoretical 
nature. It seems to be true from actual operation, as well as theory, 
that the steeper the threads the higher the efficiency. In actual 
practice there is seldom an opportunity to change the slope of the 
thread to get increased efficiency. The slope is usually settled from 
considerations of speed ratio, or available space, or some other con¬ 
dition. The usual practical problem is to take a given worm and 
worm gear, and to make out of it as efficient a device as possible. 
With hobbed gears running in oil baths, and with moderate pressures 
and speeds, the efficiency will range between 40 per cent and 70 per 
cent. The latter figure is higher than is usually attained. 

To avoid cutting and to secure high efficiency, it seems essential 
to make the worm and the gear of different materials. The worm- 
thread surfaces being in contact a greater number of times than the 
gear teeth, should evidently be of the harder material. Hence, the 
worm is usually made of steel, and the gear of cast iron, brass, or 
bronze. To save the expense of a large and heavy bronze gear, it is 
common to make a cast-iron center and bolt a bronze rim to it. 

The worm being the most liable to replacement from wear, 
it is desirable to so arrange its shaft fastening and general accessibility 
that it may be readily removed without disturbing the worm gear. 

The circular pitch of the gear and the pitch of the worm thread 
must be the same, and the practical question comes in as to the 
threads per inch possible to be cut in the lathe in the production 
of the worm thread. The pitch must satisfy this requirement; 
hence the pitch will usually be fractional, and the diameter of the 
worm gear, to give the necessary number of teeth, must be brought 
to it. While it would perhaps be desirable to keep an even diametrical 
pitch for the worm gear, yet it would be poor design to specify a worm 
gear which could not be cut in a lathe. 

The standard involute of 15°, and the standard proportions 
of teeth as given on page 132, are usually used for worm threads. 
This system requires the gear to have at least 30 teeth, for if fewer 
teeth are used the thread of the worm will interfere with the flanks 


MACHINE DESIGN 


145 


of the gear teeth. This is a mathematical relation, and there are 
methods of preventing it by a change of tooth proportions or a change 
of angle of worm thread; but as there are few instances in which less 
than 30 teeth are required, it is not deemed worth while to go into a 
lengthy discussion of this point. 

The angle of the worm embraced by the worm-gear teeth varies 
from 60° to 90°, and the general dimensions of rim are made about 
the same as for spur gears. The arms, or the web, have the same 
reasons for their size and shape. Probably web gears and cross¬ 
shaped arms are more common than oval or elliptical sections. 

Worm gears sometimes have cast teeth, but they are for the 
roughest service only, and give but a point bearing at the middle of 
the tooth. An accurately hobbed worm gear will give a bearing 
clear across the face of the tooth, and, if properly set up and cared 
for, makes a good mechanical device although admittedly of some¬ 
what low efficiency. 

Example. A tooth load of 1,200 lbs. is transmitted between 
two spur gears of 12-inch and 30-inch diameter, respectively, the 
latter gear making 100 revolutions per minute. Calculate a suitable 
pitch and face of tooth by the “Lewis” formula. 

Solution. Let a cast-iron gear be used. Speed of the gear 

(See Table VIII, 


.117 

W = SP'Fy 

Use machine cut teeth; then 

F = 3 P 1 

W = SP'ZP'y = 3 S(P l ) 2 y 


30 

= 100 X 3.1416 = 785+ ft. per minute. 

Page 132.) 

S = 3000 lbs. per sq. in. 

If the cycloidal tooth is used 

.684 

y ~ n 

Assume N = 90, then 

y = .124 = .124 - .007 = 

u 90 





146 


MACHINE DESIGN 


P l = 




1200 


3 Sy \ 3 X 3000 X .117 
Substituting the value of P l in equation 100 


✓ 1.14 = 1.07 


= 3.141 £ = 2 

1 07 


With 90 teeth the diametral pitch would be 3, and as 90 teeth 
were assumed in the above calculation it is not necessary to check 
the result any further 


PROBLEMS FOR PRACTICE 

1. Calculate proportions of a standard gear tooth of 1£ 
diametral pitch, making a rough sketch and putting the dimen¬ 
sions on it. 

2. Suppose the above tooth to be loaded at the top with 5,000 
lbs. If the face be 6 inches, calculate the fiber stress at the pitch 
line, due to bending. 

Note. In solving the next three problems use data given in the illus¬ 
trative example. 

3. Assuming a J-inch web on the 12-inch gear, calculate the 
shearing fiber stress at the outside of a hub 4 inches in diameter. 

4. Design elliptical arms for the 30-inch gear, allowing S = 

2 , 200 . 

5. Design cross-shaped arms for 30-inch gear. 


BEARINGS, BRACKETS, AND STANDS 


NOTATION—The following notation is used throughout the chapter on 

Bearings, Brackets, and Stands: 


A = Area (square inches) 
a = Distance between bolt centers 
(inches) 

b = Width of bracket base (inches) 
c = Distance of neutral axis from 
outer fiber (inches) 

D = Diameter of shaft (inches) 
d = Diameter of bolt body (inches) 
d l — Diameter at root of thread 
(inches) 

H = Horse-power 

h = Thickness of cap at center 
(inches) 


/t = Coefficient of friction (per cent) 
N = Number of revolutions per minute 
n = Number of bolts in cap 
n, = Number of bolts in bracket base 
P = Total pressure on bearing (lbs.) 
p = Pressure per square inch of pro¬ 
jected area (lbs.) 

S = Safe tensile fiber stress (lbs.) 

S B = Safe shearing fiber stress (lbs.) 
T = Total load on bolts at top of 
bracket (lbs.) 

t = Thicknessofbracketbase(inches) 
x = Distance from line of action of 







MACHINE DESIGN 


147 


7 = Moment of inertia load to any section of bracket 

L = Length of bearing (inches) (inches) 

Analysis. Machine surfaces taking weight and pressure of 
other parts in motion upon them are, in general, known as bearings. 
If the motion is rectilinear, the bearing is termed a slide, guide , or 
way, such as the cross-slide of a lathe, the cross-head guide of a steam 
engine, or the ways of a lathe bed. 

If the motion is a rotary one, like that of the spindle of a lathe 
the simple word bearing is generally used. 

In any bearing, sliding or rotary, there must be strength to 
carry the load, stiffness to distribute the pressure evenly over the 
full bearing surface, low intensity of such pressure to prevent the 
lubricant from being squeezed out and to minimize the wear, and 
sufficient radiating surface to carry away the heat generated by 
the friction of the surfaces as fast as it is generated. Sliding bearings 
are of such varied nature, and exist under conditions so peculiar to 
each case, that a general analysis is practically impossible. 

Rotary bearings can be more definitely studied, as there are 
but two variable dimensions, diameter and length, and it is the 
proper relation between these two facts that determines a good bear¬ 
ing. The size of the shaft, as noted under “Shafts,” is calculated by 
taking the bending moment at the center of the bearing, combin¬ 
ing it with the twisting moment, and solving for the diameter con¬ 
sistent with the assumed fiber stress. But this size must then be 
tried for deflection due to the bending load, in order that the require¬ 
ment for stiffness may be fulfilled. When this is accomplished, the 
friction at the bearing surface may still generate so much heat that 
the exposed surface of the bearing will not radiate it as fast as gener¬ 
ated, in which case the bearing gets hotter and hotter, until it finally 
burns out the lubricant and melts the lining of the bearing, and ruin 
results. 

The heat condition is usually the critical one, as it is very 
easy to make a short bearing which is strong enough and amply 
stiff for the load it carries, but which nevertheless is a failure as a 
bearing, because it has so small a radiating surface that it cannot 
run cool. 

The side load, which causes the friction and the consequent de¬ 
velopment of heat, is due to the jpull of the belt in the case of pulleys, 


148 


MACHINE DESIGN 


the load on the teeth of gears, the 'pull on cranks and levers, the weight 
of parts , etc. If pure torsion could be exerted on shafts without 
any side pressure, and all the weight that comes on the shaft counter¬ 
acted, trouble would not be encountered due to the development of 
heat in bearings; in fact, there would theoretically be no need of 
bearings, as the shafts would naturally spin about their axes, and 
would not need support. 

It can be shown, theoretically, that the radiating surface of a 
bearing increases relatively to the heat generated by a given side 
load, only when the length of the bearing is increased. In other 
words, increasing the diameter and not the length, theoretically 
increases the heat generated per unit of time just as much as it 
increases the radiating surface; hence nothing is gained, and heat 
accumulates in the bearing as before. This important fact is 
verified by the design of high-speed bearings, which, it is always 
noted, are very long in proportion to their diameter, thus giving 
relatively high radiating power. 

Bearings must be rigidly fastened to the body of the machine 
in some way, and the immediate support is termed a bracket , frame , 
or housing. Bracket is a very general term, and applies to the sup¬ 
ports of other machine parts besides bearings. It is especially ap¬ 
plicable to the more familiar types of bearing supports, and is here 
introduced to make the analysis complete. 

The bracket must be strong enough as a beam to take the side 
load, the bending moment being figured at such points as are neces¬ 
sary to determine its outline. It may be of solid, box, or ribbed 
form, the latter being the most economical of material, and usually 
permitting the simplest pattern. The fastening of the bracket to 
the main body of the machine must be broad to give stability; the 
bolts act partly in shear to keep the bracket from sliding along its 
base, and partly in tension to resist its tendency to rotate about 
some one of its edges, due to the side pull of the belt, gear tooth, 
or lever load, as the case may be. The weight of the bracket itself 
and of the parts it sustains through the bearing, has likewise to be 
considered; and this acts, in conjunction with the working load on 
the bearing, to modify the direction and magnitude of the resultant 
load on the bracket and its fastening. 

Stands are forms of brackets, and are subject to the same analysis. 


MACHINE DESIGN 


149 


The distinction is by no means well defined, although a stand is usually 
considered as having an upright or inverted position with reference 
to the ground. The ordinary “hanger” is a good example of an in¬ 
verted stand; and the regular “floor stand,” found on jack shafts 
in some power houses, is an example of the general class. 

Theory. As the method of calculation of the diameter of the 
shaft, as well as its deflection, has been considered under “Shafts,” 
the theoretical study of bearings 
may be assumed as starting on 
a given basis of shaft diameter 
D. The main problem then 
being one of heat control, let 
the amount of heat developed 
in a bearing by a given side load 
be first calculated. The force of 
friction acts at the circumference 
of the shaft, and is equal to 
the coefficient of friction times 
the normal force; or, for a 
given side load P, Fig. 84, the 
force of friction would be jiP. The peripheral speed of the 

•rrDN 

shaft for N revolutions per minute is ——— feet per minute. As work 

is force times distance, the work wasted in friction is thenfoot- 
J 12 

pounds per minute. One horse-power being equal to 33,000 foot¬ 
pounds per minute 

12 X 33000 ^ } 



The value of for ordinary, well-lubricated bearings, may run as 
low as 5 per cent; but as the lubrication is often impaired, it quite 
commonly rises to 10 or 12 per cent. A value of 8 per cent is a fair 
average. This amount of horse-power is dissipated through the 
bearing in the form of heat. If the ability of each particle of the 
metal around the shaft to transmit the heat, or to pass it along to 
the outside of the casting could be determined and then if the ability 
of the particles of air surrounding the casting to receive and carry 
away this heat could be determined, just such proportions of the 















150 


MACHINE DESIGN 


bearing and its casing as would never choke or retard this free trans¬ 
fer of heat from the running surface could be calculated. 

Such refined theory is not practical, owing to the complicated 
shapes and conditions surrounding the bearing. The best that can 
be said is that for the usual proportions of bearings the side load 
may exist up to a certain intensity of “pressure per square inch of 
projected area” of bearing, or, in form of an equation, 

P = pLD (118) 

The constant p is of a variable nature, depending on lubrication, 
speed, air contact, and other special conditions. For ordinary 
bearings having continuous pressure in one direction, and only fair 
lubrication, 400 to 500 is an average value. When the pressure 




•t L 


Hf^« 


L 


Fig. 85. Hanger 

changes direction at every half-revolution, the lubricant has a better 
chance to work fully over the bearing surface, and a higher value 
is permissible, say, 500 to 800. In locations where mere oscillation 
takes place, not continuous rotation, and reversal of pressure occurs, 
as on the cross-head pin of a steam engine, p may run as high as 
900 to 1,200. On the crank pins of locomotives, which have the 
reversal of pressure, and the benefit of high velocity through the air 
to facilitate cooling, the pressures may run equally high. On the 
eccentric crank pins of punching and shearing machines, where the 




















































MACHINE DESIGN 


151 


pressure acts only for a brief instant and at intervals, the pressure 
ranges still higher without any dangerous heating action. 

When a bearing, for practical reasons, is provided with a cap 
held in place by bolts or studs, the theory of the cap and bolts is of 
little importance, unless the load comes directly against the cap and 
bolts. Except in the latter case, the proportions of the cap and the 
size of the bolts are dependent upon general appearance and 
utility, it being manifestly desirable to provide a substantial design, 
even though some excess of strength is thereby introduced. 

For the worst case of loading, however, which is when the 
cap is acted upon by the direct load, such as P in Fig. 85, there exists 
a condition of a centrally loaded beam supported at the bolts. It 
is probable that the beam is partially fixed at the ends by the clamp¬ 
ing of the nut; also that the load P, instead of being concentrated 
at the center, is to some extent distributed. It is hardly fair to 

i i , Pa Pa . . . 

assume the external moment equal to -tt or —, the one being too 

o 4 


small, perhaps, and the other too large. It will be reasonable to 

p a 

take the external moment at —* in which case, equating the external 
moment to the internal moment of resistance, 


Pa SI SLh 2 
6 c 6 


(119) 


from which, the length of bearing being known, the thickness h may 
be calculated. 

One bolt on each side is sufficient for bearings not more than 
6 inches long, but for longer bearings usually two bolts on a side are 
used. The theoretical location for two bolts on a side, in order 
that the bearing may be equally strong at the bolts and at the center 
of the length, may be shown by the principles of mechanics to be 

—L from each end, as indicated in Fig. 85. 

24 

The bolts are evidently in direct tension, and if equally loaded 
would each take their fractional share of the whole load P. This 

2 

is difficult to guarantee, and it is safer to consider that — P may be 
taken by the bolts on one side. On this basis, assuming a total num- 







152 


MACHINE DESIGN 


ber of bolts ft, 
of the bolts 


and equaling the external force to the internal resistance 



( 120 ) 


from which the proper commercial diameter may be readily found. 

The bracket may have the shape shown in Fig. 86. The por¬ 
tion at B is under direct shearing stress; and if A be the area at the 




point, and S a the safe shearing stress, then, equating the external 
force to the internal shearing resistance, 

P = AS 9 (121) 

The same shear comes on all parts of the bracket to the left of the 
load, but there is an excess of shearing strength at these points. 

At the point of fastening, the bolts are in shear, due to the same 
load, for which the equation is 


7 T(P 

P — — T-n t S c 


(122) 



















































































MACHINE DESIGN 


153 


For the upper bolts, the case is that of direct tension, assuming 
that the whole bracket tends to rotate about the lower edge E. To 
find the load T on these bolts, moments about the point E are taken 
as follows: 

PL, = Tl 
or 



(123) 


Then, equating the external force to the internal resistance, 



ircL 


X 



(124) 


The upper flange is loaded with the bolt load T, and tends to 
break off at the point of connection to the main body of the bracket, 
the external moment, therefore, being Tr. The section of the 
flange is rectangular; hence the equation of external and internal 
moments is 

rp PL, Sbf 

lr = * j ' r = — 



It may be noted that the lower bolts act on such a small leverage 
about E , that they would stretch and thus permit all the load to be 
thrown on the upper bolts; this is the reason why they are not subject 
to calculation for tension. 

The section of the bracket to the left of the load P is dependent 
upon the bending moment, for, if this section is large enough to take 
the bending moment properly, the shear may be disregarded. It 
should be calculated at several points, to make sure that the fiber 
stress is within allowable limits. The general expression for the 
equation of moments is, for any section at leverage x, 



from which, by the proper substitution of the moment of inertia of 
the section, the fiber stress may be calculated. The moment of 
inertia for simple ribbed sections may be found in most handbooks. 
The process of solution of the above equation, though simple, is 
apt to be tedious, and is not considered necessary. 








154 


MACHINE DESIGN 


Practical Modification. Adjustment. Adjustment is an im¬ 
portant practical feature of bearings. Unless the proportions are 
so ample that wear is inappreciable, simple and ready adjustment 
must be provided. The taper bushing, Fig. 87, is neat and satisfac- 




Fig. 87. Taper Bushing 

tory for machinery in which expense and refinement are permissible. 
Though this is true of some machine tools, it is not true of the general 
run of bearings. The most common form of adjustment is secured 

by the plain cap—which may or may not be 
tongued into the bracket—with liners 
placed in the joint when new, which may 
subsequently be removed or reduced so as 
to allow the cap to close down upon the 
shaft. Several forms of cap bearings are 
illustrated in Figs. 88, 89, and 90. 

Large engine shaft bearings have special 
forms of adjustment such as wedges and 
screws, which take up the wear in all 
directions at the same time accurately 
preserving the alignment of the shafts; but this refinement is seldom 
required for shafts of ordinary machinery. 

In cases where the cap bearing is not applicable, a simple bush¬ 
ing may be used. This may be removed when worn, and a new one 
inserted, the exact alignment being maintained, as the outside will 



Fig. 88. Tongued Cap Bearing 















































MACHINE DESIGN 


155 


be concentric with the original axis of shaft, regardless of the wear 
which has taken place in the bore. 

Lubrication. The lubrication of bearings is a part of the design, 
in that the lubricant should be introduced at the proper point, and 
pains taken to guarantee its distribution to all points of the running 
surface. The method of lubrication should be so certain that no 
excuse for its failure would be possible. Grease is a successful lubri¬ 
cator for heavy loads and slow speeds, oil for light loads and high 
speeds. 

In order to insure the lubricant reaching the sliding surfaces 
and entering between them, it must be introduced at a point where 
the pressure is moderate, and where the motion of the parts will 




Fig. 89. Plain Cap Bearing Fig. 90. Bracket Forming Tongue 


naturally lead it to all points of the bearing. Grooves and channels 
of ample size assist in this regard. A special form of bearing uses 
a ring riding on the shaft to carry the oil constantly from a small 
reservoir beneath the shaft up to the top, where it is distributed along 
the bearing and finally flows back to the reservoir and is used again. 

Materials. The materials of which bearings are made vary 
with the service required and with the refinement of the bearing. 
Cast iron makes an excellent bearing for light loads and slow speeds, 
but it is very apt to “seize” the shaft in case the lubrication is in the 
least degree impaired. Bronze, in its many forms of density and 
hardness, is extensively used for high-grade bearings, but it also 
has little natural lubricating power, and requires careful attention 
































156 


MACHINE DESIGN 


to keep it in good condition. Babbitt, a composition metal of vary¬ 
ing degrees of hardness, is the most universal and satisfactory ma¬ 
terial for ordinary bearings. It affords a cheap method of produc¬ 
tion, being poured in molten form around a mandrel, and firmly 
retained in its casing or shell through dovetailed pockets into which 
the metal flows and hardens. It requires no boring or extensive 
fitting; some scraping to uniform bearing is necessary in most cases, 
but this is easily and cheaply done. Babbitt is a durable material, 
and has some natural lubricating power, so that it has less tendency 
to heat with scanty lubrication than any of the materials previously 
mentioned. Almost any grade of bearing may be produced with 
babbitt. In its finest form the babbitt is hammered, or pened, 
into the shell of the bearing, and then bored out nearly to size, a 
slightly tapered mandrel being subsequently drawn through, com¬ 
pressing the babbitt and giving a polished surface. 

A combination bearing of babbitt and bronze is sometimes 
used. In this the bronze lies in strips from end to end of the bearing, 
and the babbitt fills in between the strips. The shell, being of bronze, 
gives the required stiffness, and the babbitt the favorable running 
quality. 

Examples. 1. The journals on the tender of a locomotive are 
3J X 7 inches. The total weight of the tender and load is 60,000 
pounds. If there are 8 journals, what is the pressure per square 
inch of the projected area? 

Solution. Projected area of one journal = 3J X 7 = 24J sq. in. 
Total projected area = 8 X 24J = 196 sq. in 


60000 

196 


306 lbs. per sq. in. 


2. What horse-power is lost in friction at the circumference 
of a 3-inch bearing carrying a load of 6,000 pounds, if the number 
of revolutions per minute is 150 and the coefficient of friction is 
assumed to be 5 per cent? 

Solution. 

_ iiPttDN 
~ 12 X 33000 

TJ .05 X 6000 X 3.1416 X 3 X 150 

= 1.07 


12 X 33000 





MACHINE DESIGN 


157 


PROBLEMS FOR PRACTICE 

1. The allowable pressure on a bearing is 300 pounds per 
square inch of projected area. What is the required length of the 
bearing if the total load is 4,500 pounds and the diameter is 3 inches? 

2. The cross-head pin of a steam engine must be 2.5 inches 
in diameter to withstand the shearing strain. If the maximum 
pressure is 10,000 pounds, what length should be given to the pin? 

3. The cast-iron bracket in Fig. 86 has a load P of 1,000 pounds. 
Determine the fiber stress in the web section at the base of the bracket 
if the thickness is taken at ^ inch, and 1^= 12 inches; / = 20 inches; 
k = 11 inches. 

4. Calculate the diameter of the bolts at the top of the bracket. 

5. Assuming r equal to 6 inches, what is the fiber stress at the 
root of flange? 


INDEX 


INDEX 


^ PAGE 

Alloy steels. 43 

Alloys. 45 

Aluminum. 44 

American system of rope driving. 96 

B 

Beams. 13, 22 

Bearing alloys. 45 

Bearing value of plate. 58 

Bearings, brackets, and stands. 146 

analysis. 147 

practical modification... 154 

theory. 149 

Belts, leather. 82 

Bending moment. 19 

Bevel gears. 138 

analysis. 138 

practical modification. 139 

theory. 138 

Bolts, studs, nuts, and screws. 45 

analysis. 46 

practical modification. 52 

theory. 48 

Brackets. 146 

Brass. 44 

Breaking strength. 11 

Bronze. 44 

Butt joint. 57 

C 

Calculations, notes, and records. 6 

Calking. 60 

Cantilever beam. 13, 103 

Cap screws. 55 

Cast iron. 40 

Centrifugal whirling. 120 

Circular pitchy. 128 

Clamp coupling. 75 

Claw coupling. 75 

Combined stresses. 28 

Compression. 12 

Compressive stress. 8 

Continuous system of rope driving. 96 

Copper. 43 










































2 INDEX 

PAGE 

Cotters. 67 

analysis. 67 

practical modification. 70 

theory..*. 68 

Cotton rope. 93 

Couplings. 72 

analysis. 72 

practical modification. 74 

theory. 73 

Cover plates. 57 

Cross-hatching. 35 

Cylinders, strength of. 29 

D 

Deflection of beams. 23 

Deflection of shaft. 118 

Deformation. 9 

Delineation and specification. 34 

detail drawings. 35 

general drawing. 36 

Design, method of.;. 30 

analysis of conditions and forces. 31 

delineation and specification. 34 

practical modification. 32 

theoretical design. 31 

Diametral pitch. 128 

Drawings. 35 

E 

Efficiency of joint. 59 

Elasticity. 10 

Empirical data. 5 

English system of rope driving. 96 

External shear. 19 

F 

Factor of safety. H 

Fatigue. 39 

Feather keys. 65 

Fiber stresses. 19 

Flange coupling. 72 

Friction clutches. 77 

analysis. 73 

practical modification. 81 

theory.. 79 

Full-shrouding. I 35 

Fusible alloys. 45 











































INDEX 


* 


3 


Gears 


G 


PAGE 

126, 138, 140 


H 


Half-shrouding. 135 

Handbooks. 5 

Hemp and cotton rope. 93 

analysis... 93 

materials. 95 

practical modification. 94 

systems of rope driving. 99 

theory. 93 

Hob, definition of. 141 

Hook-tooth gear. 136 

Hooke’s law. 119 

Horsepower of shafting.:. 120 


I 

Invention. 5 

J 

Joint. 57, 59 

efficiency of. 59 

kinds of. 57 


K 

Keys and pins. 62 

analysis. 62 

practical modification. 65 

theory. 63 

L 

Lap joint.... 57 

Laws of stress. 9 

Leather belts. 82 

analysis. 83 

care of. 90 

initial tension in. 89 

material of. 89 

methods for joining ends of. 91 

practical modification. 86 

speed of. 88 

strength of. 88 

theory. 85 

Lewis formula. 132 

Lock nuts. 53 

Lubrication.:. 37 

friction. 37 

lubricant. 35 

means for lubrication. 1 . 38 









































4 


INDEX 


M PAGE 

Machine design. 1-157 

bearings, brackets, and stands. 146 

bevel gears. 138 

bolts, studs, nuts, and screws. 45 

couplings. 72 

friction clutches. 77 

introduction. 1 

keys, pins, and cotters. 62 

leather belts. 82 

lubrication. 37 

method of design. 30 

pulleys. 100 

rivets and riveted joints. 57 

rope gearing. 93 

shafts. Ill 

spur‘gears. 126 

strength of materials. 8 

worm and worm gear. 140 

Machine screws. 56 

Main plates. 58 

Malleable cast iron. 42 

Materials, strength of. 8 

Materials employed in construction. 39 

aluminum. 44 

bearing alloys. 45 

brass. 44 

bronze. 44 

cast iron. 40 

copper. 43 

fusible alloys. 45 

steel. 43 

wood. 45 

wrought iron. 42 

Moment of a force. 14 

Mortise teeth. 135 

Muff coupling. 74 

Multiple system of rope driving. 96 

N 

Net section. 58 

Neutral axis. 22 

Neutral surface. 22 

Nuts. 45 

O 

Oldham coupling. 76 

• « 

P 

Permanent set. 10 

Pipes and cylinders, strength of. 29 














































INDEX 


5 


PAGE 

Pitch. 58 

Pitch cylinders. 127 

Principle of moments. 15 

Pulleys. 100 

analysis. 100 

practical modification. 106 

theory. 102 

R 

Rankine formula. 120 

Reactions on beams. 16 

Resisting moment. 22, 26 

Rim, arms, and hub of spur gears. 133 

analysis. 133 

practical modification. 134 

theory. 133 

Rivets and riveted joints. 57 

analysis. 57 

practical modification. 59 

theory. 58 

Rope driving, systems of. 96 

Rope gearing. 93 

S 

Screws. 45 

Section modulus. 22 

Set screws. 56 

Shafts. 24, 111 

analysis. Ill 

function of. 24 

practical modification. 121 

strength of. 24 

theory. 114 

Shear. 13 

Shearing stress. 8, 13 

Shearing value of rivet. 58 

Simple beam. 13 

Splines. 63 

Split pulleys. 108 

Spur gears. 126 

analysis. 127 

rim, arms, and hub. 133 

Stands. 146 

Steel. 43 

Strain. 9 

Strength of beams. 22 

deflections of beams. 23 

resisting moment of beams. 22 














































6 


INDEX 


PAGE 


Strength of materials. 8 

bending moment. 19 

combined stresses. 28 

external shear. 19 

strength of beams. 22 

strength of pipes and cylinders. 29 

strength of shafts. 24 

stress. 8 

Strength of pipes and cylinders. 29 

Strength of shafts. 24 

formula for. 27 

resisting moment. 26 

shafts. 24 

torsional stress. 26 

twisting moment. 24 

Stress. 8 

beams. 13 

compression. 12 

elasticity. 10 

experimental laws. 9 

factor of safety. 11 

moment of a force. 14 

principle of moments. 15 

reactions on beams. 16 

safe working strength. 11 

shear. 13 

summation of forces. 14 

tension. 12 

testing machines. 12 

ultimate stress. 11 

unit deformation. 9 

unit stress. 8 

Stub tooth. 136 

Studs. 45, 55 

Summation of forces. 14 


T 

Tables 

factor of safety. 11 

gear design data. 137 

physical constants, formulas for. 23 

physical constants graphically defined. 21 

proportions for feather keys. 67 

proportions for gib keys. 67 

safe working stresses for different speeds. 132 

sizes of leather belting. 90 

strength of bolts. 50 

Tap bolts. 55 

Tensile stress. 8 
















































INDEX 


7 


PAGES. 

Tension... 12 

Theoretical design. 31 

Through bolts. 55 

Tightening pulley. 87 

Torsional stress. 26 

Twisting moment. 24 

U 

Ultimate stress. 11 

Unit deformation. 9 

Unit stress. 8 

Universal coupling. 76 

W 

Wire rope. 97 

practical modification.:. 97 

pulleys for. 98 

theory. 97 

Wood. 45 

Woodruff key. 66 

Working strength. 11 

Worm and worm gear. 140 

analysis. 141 

practical modification. 144 

theory. 141 

Wrought iron. 42 

























\ 

















